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NEW 
ANALYTIC  GEOMETRY 


BY 

PERCEY  F.  SMITH,  Ph.D. 

PROFESSOR  OF  MATHEMATICS  IN  THE  SHEFFIELD  SCIENTIFIC  SCHOOL 
OF   YALE    UNIVERSITY 


ARTHUR  SULLIVAN  GALE,  Ph.D. 

PROFESSOR  OF  MATHEMATICS   IN  THE    UNIVERSITY   OF   ROCHESTER 


GINN  AND  COMPANY 

HOSTON     •     NEW    YORK     •     CHICAGO     •     LONDON 
ATLANTA      •     DALLAS     •     COLUMBUS     •     SAN    FRANCISCO 


COPYRIGHT,  1904,  1905,  BY 

ARTHUR  SULLIVAN  GALE 


COPYRIGHT,  1912,   BY 
PERCEY  F.  SMITH 


ALL  RIGHTS  RESFRVED 
919.2 


)Cl)e    StbenKum    •jj^vtss 

GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

A  glance  at  the  Table  of  Contents  of  the  present  volume 
will  reveal  the  fact  that  the  subject  matter  differs  in  many 
respects  from  that  included  in  current  textbooks  on  analytic 
geometry.  The  authors  have  recognized  the  great  importance, 
in  the  applications,  of  the  exponential  and  trigonometric  func- 
tions, of  "  setting  up  "  and  studying  functions  by  their  graphs, 
of  parametric  equations  and  the  locus  problem,  and  of  "fitting" 
curves  to  points  determined  by  empirically  given  data.  To  meet 
this  need  chapters  have  been  included  covering  all  these  topics. 
The  discussion  in  Chapter  VI  of  transcendental  curves  and 
equations  is  intended  to  be  thorough,  and  tables  are  provided, 
whenever  useful,  to  lighten  the  labor  of  computation.  A  stu- 
dent loses  interest  in  a  function  if  he  is  unable  to  calculate 
rapidly  its  numerical  values.  The  problems  of  Chapter  VIII 
provide  a  large  variety  of  functions  arising  from  applied  prob- 
lems, and  careful  "  graphing "  and  measurement  of  maximum 
ind  minimum  values  are  emphasized.  The  text  of  Chapter  XII 
3n  Parametric  Equations  and  Loci  is  unusually  complete,  and 
^are  is  taken  to  familiarize  the  student  with  those  curves  which 
)ccur  in  applied  mathematics.  The  study  of  locus  problems 
)y  means  of  parametric  equations  is  amply  illustrated.  Chap- 
ier  XX  presents  the  topic  of  empirical  equations  and  contains 
I,  wide  variety  of  problems. 

The  authors  have  not  neglected  to  provide  an  adequate  and 
horough  drill  in  the  use  of  coordinates  and  in  the  employ- 
aent  of  analytic  methods.  It  is  acknowledged  that  this  is  the 
irimary  aim  of  analytic  geometry.    The  proofs  will  be  found 

lii 


iv  NEW  ANALYTIC  GEOMETRY 

simple  and  direct.  The  chapters  devoted  to  the  study  of  the 
conic  sections  (Chapters  X  and  XI)  are  brief  but  contain  all 
essential  characteristics  of  these  important  curves.  The  ex- 
amples are  numerous,  and  many  are  given  without  answers  in 
case  any  useful  purpose  is  served  by  so  doing.  The  book,  like 
the  authors'  "  Elements  of  Analytic  Geometry,"  is  essentially  a 
drill  book ;  but  at  the  same  time  all  difiiculties  are  not  smoothed 
out,  though  the  student  is  aided  in  making  his  own  way.  He 
is  taught  to  formulate  rules  descriptive  of  methods,  and  to 
summarize  the  main  results.  Tlie  appearance  in  the  text  of 
various  Rules  is  designed  expressly  to  encourage  the  student 
in  the  habit  of  formulating  precise  statements  and  of  making 
clear  to  himself  each  new  acquisition. 

Acknowledgments  are  due  to  Dr.  George  F.  Gundelfinger, 
of  the  department  of  mathematics  of  the  Sheffield  Scientific 
School,  for  many  of  the  problems  in  the  analytic  geometry  of 
space  and  for  many  valuable  suggestions. 

THE  AUTHORS 

Nkw  ITavkn,  Connecticut 


CONTENTS 


CHAPTER  I 

FORMULAS  AND  TABLES  FOR  REFERENCE 
SECTION  PAGE 

1.  Formulas  from  geometry,  algebra,  and  trigonometry 1 

2.  Three-place  table  of  common  logarithms  of  numbers 4 

3.  Squares  and  cubes  ;  square  roots  and  cube  roots 5 

4.  Natural  values  of  trigonometric  functions 6 

5.  Logarithms  of  trigonometric  functions 6 

6.  Natural  values.    Special  angles 7 

7.  Rules  for  signs  of  trigonometric  functions 7 

CHAPTER  II 
CARTESIAN  COORDINATES 

8.  Directed  line 8 

9.  Cartesian  coordinates 9 

10.  Rectangular  coordinates 10 

11.  Lengths 13 

12.  Inclination  and  slope 1(5 

13.  Point  of  division 19 

14.  Areas 24 


CHAPTER  III 
CURVE  AND  EQUATION 

15.  Locus  of  a  point  satisfying  a  given  condition 30 

16.  Equation  of  the  locus  of  a  point  satisfying  a  given  condition  .    .  30 

17.  First  fundamental  problem 32 

18.  Locus  of  an  equation 36 

19.  Second  fundamental  problem 37 

20.  Third  fundamental  problem.     Discussion  of  an  equation     ...  42 

V  * 


vi  NEW  ANALYTIC  GEOMETRY 

SECTION  PAGE 

21.  Directions  for  discussing  an  equation 47 

Sign  of  a  quadratic 49 

22.  Asymptotes 51 

23.  Points  of  intersection 55 


CHAPTER  IV 
THE  STRAIGHT  LmE 

24.  The  degree  of  tlie  equation  of  any  straight  line 58 

25.  Locus  of  any  equation  of  the  first  degree 59 

26.  Plotting  straight  lines 60 

27.  Point-slope  form 62 

28.  Two-point  form 63 

29.  Intercept  form 67 

30.  Condition  that  three  lines  shall  intersect  in  a  connnon  point    .    .  67 

31.  Theorems  on  projection 68 

32.  The  normal  equation  of  the  straight  line 70 

33.  Reduction  to  the  normal  form 71 

34.  The  perpendicular  distance  from  a  line  to  a  point 74 

35.  The  angle  which  a  line  makes  with  a  second  line 80 

36.  Systems  of  straight  lines 84 

37.  System  of  lines  passing  through  the  intersection  of  two  given  lines  87 

CHAPTER  V 
THE  CIRCLE 

38.  Equation  of  the  circle 92 

39.  Circles  determined  by  three  conditions 94 

CHAPTER  VI 
TRANSCENDENTAL  CURVES  AND  EQUATIONS 

40.  Natural  logarithms 101 

Table  of  values  of  the  exponential  function 104 

41.  Sine  curves 105 

42.  Addition  of  ordinates •  .    .    .  Ill 

43.  Boundary  curves 114 

44.  Transcendental  equations.    Graphical  solution 116 


CONTENTS  vii 
CHAPTER  VII 

POLAR  COORDINATES 
SECTION  PAGE 

45.  Polar  coordinates 119 

46.  Locus  of  an  equation 120 

47.  Rapid  plotting  of  polar  equations 125 

48.  Points  of  intersection 127 

49.  Transformation  from  rectangular  to  polar  coordinates  ....  128 

50.  Applications.    Straight  line  and  circle 130 


CHAPTER  VIII 
FUNCTIONS  AND  GRAPHS 

51.  Functions 134 

52.  Notation  of  functions 143 

CHAPTER  IX 
TRANSFORMATION  OF  COORDINATES 

53.  Introduction 144 

54.  Translation  of  the  axes 144 

55.  Rotation  of  the  axes 146 

56.  General  transformation  of  coordinates 148 

57.  Classification  of  loci 148 

58.  Simplification  of  equations  by  transformation  of  coordinates     .  149 

CHAPTER  X 
PARABOLA,  ELLIPSE,  AND  HYPERBOLA 

59.  The  parabola 153 

60.  Construction  of  the  parabola 157 

61.  Theorem 159 

62.  The  ellipse 159 

63.  Construction  of  the  ellipse 163 

64.  Theorem 165 

65.  The  hyperbola 165 

66.  Conjugate  hyperbolas  and  asymptotes 170 

67.  Equilateral  or  rectangular  hyperbola 173 


viii  NEW  ANALYTIC  GEOMETRY 

SECTION  PAGE 

68.  Construction  of  the  hyperbola 173 

69.  Theorem 175 

70.  Locus  of  any  equation  of  the  second  degree 176 

71.  Plotting  the  locus  of  an  equation  of  the  second  degree  ....  180 

72.  Conic  sections 185 

Systems  of  conies 188 


CHAPTER  XI 
TANGENTS 

73.  Equation  of  the  tangent 190 

74.  Theorem 194 

75.  Equation  of  the  normal 195 

76.  Subtangent  and  subnormal 196 

77.  Tangent  whose  slope  is  given 198 

78.  Formulas  for  tangents  when  the  slope  is  given 200 

79.  Properties  of  tangents  and  normals  to  conies 201 

CHAPTER  XII 
PARAMETRIC  EQUATIONS  AND  LOCI 

80.  Plotting  parametric  equations 205 

81.  Various  parametric  equations  for  the  same  curve 208 

82.  Locus  problems  solved  by  parametric  equations 211 

83.  Loci  derived  by  a  construction  from  a  given  curve 218 

84.  Loci  using  polar  coordinates 220 

85.  Loci  defined  by  the  points  of  intersection  of  systems  of  lines     .  223 
Diameters  of  conies 226 

CHAPTER  XIII 
CARTESIAN  COORDINATES  IN  SPACE 

86.  Cartesian  coordinates 230 

87.  Orthogonal  projections 233 

88.  Direction  cosines  of  a  line 236 

89.  Lengths 238 

90.  Angle  between  two  directed  lines 240 

91.  Point  of  division 242 


CONTENTS  ix 

CHAPTER  XIV 

SURFACES,  CURVES,  AND  EQUATIONS 
SECTION  PAGE 

92.  Loci  in  space 245 

93.  Equation  of  a  surface.    First  fundamental  problem  .    .    .    .    .  245 

94.  Planes  parallel  to  the  coordinate  planes 249 

95.  Equations  of  a  curve.   Eirst  fundamental  problem 249 

96.  Locus  of  one  equation.    Second  fundamental  problem      .    .    .  253 

97.  Locus  of  two  equations.    Second  fundamental  problem    .    .    .  253 

98.  Discussion  of  the  equations  of  a  curve.    Third  fundamental 

problem 253 

99.  Discussion  of  the  equation  of  a  surface.    Third  fundamental 
problem 256 

CHAPTER  XV 
THE  PLANE 

100.  The  normal  form  of  the  e(iuation  of  tiie  plane 260 

101.  The  general  equation  of  the  first  degree 261 

102.  Planes  determined  by  three  conditions 266 

103.  The  equation  of  a  plane  in  terms  of  its  intercepts 269 

104.  The  perpendicular  distance  from  a  plane  to  a  point      ....  269 

105.  The  angle  between  two  planes 271 

106.  Systems  of  planes 273 

CHAPTER  XVI 
THE  STRAIGHT  LINE  IN  SPACE 

107.  General  equations  of  the  straight  line 277 

108.  The  projecting  planes  of  a  line 280 

lOP.  Various  forms  of  the  equations  of  a  line 282 

110.  Relative  positions  of  a  line  and  plane        287 

CHAPTER  XVII 
SPECIAL  SURFACES 

111.  Introduction 291 

112.  The  sphere 291 

113.  Cylinders 295 

114.  The  projecting  cylinders  of  a  curve 297 


X  NEW  ANALYTIC  GEOMETRY 

SECTION  PAGE 

115.  Parametric  equations  of  curves  in  space 300 

116.  Cones 302 

117.  Surfaces  of  revolution 304 

118.  Ruled  surfaces 307 


CHAPTER  XVIII 

TRANSFORMATION  OF  COORDINATES.    DIFFERENT  SYSTEMS 
OF  COORDINATES 

119.  Translation  of  the  axes 310 

120.  Rotation  of  the  axes 310 

121.  Polar  coordinates 313 

122.  Spherical  coordinates 313 

123.  Cylindrical  coordinates 314 

CHAPTER  XIX 

QUADRIC   SURFACES  AND  EQUATIONS  OF   THE   SECOND   DEGREE 
IN  THREE  VARIABLES 

124.  Quadric  surfaces 310 

125.  Simplification  of  the  general  equation  of  the  second  degree  in 

three  variables 317 

126.  The  ellipsoid 319 

127.  The  hyperboloid  of  one  sheet 320 

128.  The  hyperboloid  of  two  sheets 321 

129.  The  elliptic  paraboloid 324 

130.  The  hyperbolic  paraboloid 325 

131.  Rectilinear  generators 327 

CHAPTER  XX 

EMPIRICAL  EQUATIONS 

132.  Introduction 330 

1.33.  Straight-line  law 330 

134.  Laws  reduced  to  straight-line  laws 333 

135.  Miscellaneous  laws 338 

136.  Conclusion 340 

INDEX 341 


IfEW  ANALYTIC  GEOMETRY 

CHAPTER  I 

FORMULAS  AND  TABLES  FOR  REFERENCE 

1.  Occasion  will  arise  in  later  chapters  to  make  use  of  the 
following  formulas  and  theorems  proved  in  geometry,  algebra, 
and  trigonometry. 

1.  Circumference  of  circle  =  2  7rr.* 

2.  Area  of  circle  =  ■jtr'^. 

3.  Volume  of  prism  =  Ba. 

4.  Volume  of  pyramid  =  i  Ba. 

5.  Volume  of  right  circular  cylinder  =  7rr%. 

6.  Lateral  surface  of  right  circular  cylinder  =  2irra. 

7.  Total  surface  of  right  circular  cylinder  =  2irr{r  +  a). 

8.  Volume  of  right  circular  cone  =  \  irr"a. 

9.  Lateral  surface  of  right  circular  cone  =  Trrs. 

10.  Total  surface  of  right  circular  cone  =  7rr{r  +  s). 

11.  Volume  of  sphere  =  -f  Trr^. 

12.  Surface  of  sphere  =  47rr'^. 

13.  In  a  geometrical  series, 

rl  —  a      a(Y"  —  1) 
r  —  1  r  —  1 

a  —  first  term,  r  =  common  ratio,  I  =  nth  term,  s  =  sum  of  n  terms. 

14.  log a6  =  logo  4- log 6.      17.  log -v/a  = -loga.      19.  logaa  =  l. 

15.  log- =  log  a  — log  &.        18.  logl=0.  20.  log- =  — log  a. 

b  1 

16.  log  a"  =  nloga. 

*  In  formulas  1-12,  r  denotes  radius,  a  altitude,  B  area  of  l)ase,  and  .s  slant 

lieiglit. 

1 


NEW  ANALYTIC  GEOMETRY 


Functions  of  an  angle  in  a  right  triangle.    In  any  right  triangle  one  of 
whose  acute  angles  is  A,  the  functions  of  A  are  defined  as  follows  : 


21. 


sin  J.  = 


cos  J.  = 


tan^  = 


opposite  side 
hypotenuse 

adjacent  side 
hypotenuse 

opposite  side 
adjacent  side 


CSC  A 


sec  A  = 


cot  J. 


hypotenuse 
opposite  side 

hypotenuse 
adjacent  side 

adjacent  side 
opposite  side 


From  the  above  the  theorem  is  easily  derived  : 

22.  In  a  right  triangle  a  side  is  equal  to  the  product  of  the  hypote- 
nuse and  the  sine  of  the  angle  opposite  to  that  side,  or  to  the  product  of 
the  hypotenuse  and  the  cosine  of  the  angle  adjacent 
to  that  side. 

Angles  in  general.  In  trigonometry  an  angle  JlOA 
is  considered  as  generated  by  the  line  OA  rotating 
from  an  initial  position  OX.  The  angle  is  positive 
when  OA  rotates  from  OJT  counter-clockwise,  and 
negative  when  this  direction  of  rotation  of  OA  is 
clockwise. 

The  fixed  line  OX  is  called  the  initial  line,  the  line  OA  the  terminal  line. 

Measurement  of  angles.  There  are  two 
important  methods  of  measuring  angular 
magnitude  ;  that  is,  there  are  two  unit 
angles. 

Degree  measure.  The  unit  angle  is  3^^  of 
a  complete  revolution,  and  is  caMed  a,  degree. 

Circular  measure.  The  unit  angle  is  an 
angle  whose  subtending  arc  is  equal  to  the 
radius  of  that  arc,  and  is  called  a  radian. 

The  fundamental  relation  between  the  unit  angles  is  given  by  the 
equation 

23.  180  degrees  =  tt  radians  (tt  =  .3.141.59  ••  •). 

Or  also,  by  solving  this, 


24. 

25. 


1  degree  == =  .0174  •  •  •  radians. 

180 

1  QQ 

1  radian  =  —  =  .57.20  ■  •  ■  degrees. 


FORMULAS  AND  TABLES  FOR  REFERENCE  3 

These  equations  enable  us  to  change  from  one  measurement  to  another. 
In  the  higher  mathematics  circular  measure  is  always  used,  and  will  be 
adopted  in  this  book. 

The  generating  line  is  conceived  of  as  rotating  around  O  through  as 
many  revolutions  as  we  choose.    Hence  the  Important  result : 

Any  real  number  is  the  circular  measure  of  some  angle,  and  conversehj, 
any  angle  is  measured  by  a  real  number. 

«o  1  1  1 

26.  cot  X  = ;  sec  x  = ;  esc  x  = 

tan  X  cos  x  sin  x 

-„    ^  slnx  cosx 

27.  tan  x  = ;  cot  x  — 

cos  x  sin  X 

28.  sin^x  +  cos-x  =  1 ;  1  +  tan^x  =  sec-x ;  1  +  cot^x  =  csc^x. 

29.  sin  (—  x)  =  —  sin  x  ;  esc  (—  x)  =  —  esc  x  ; 
cos  (—  x)  =  cos  X  ;  sec  (—  x)  =  sec  x  ; 
tan(—  x)  =  —  tanx  ;  cot(—  x)  =—  cotx. 

30.  sin  (tt  —  x)  =  sin  x  ;  sin  (tt  +  x)  =  —  sin  x  ; 
cos  (tt  —  x)  =  —  cos  X  ;  cos  (tt  +  x)  =  —  cos  x  ; 
tan  (tt  —  x)  =  —  tan  x  ;  tan  (7r  +  x)  =      tan  x. 

81.  sin  I X  I  =  cosx  ;  sin  ( — h  x  )  =      cosx  ; 

\2         /  '         \2        / 

cos  ( x)=slnx:  cos  ( — t-x|=— sinx; 

\2         /  '         \2        / 

tan  ( X  I  =  cot  X  ;  tan  (  -  +  x )  =  —  cot  x. 

\2         /  '         \2         / 

32.  sin  (2  TT  —  x)  =  sin  (—  x)  =  —  sin  x,  etc. 

33.  .sin  (x  +  ?/)  =  sin  x  cos?/  +  cosx  .sin  y. 

34.  sin  (x  —  2/)  =  sin  x  cos  y  —  cos  x  sin  y. 

35.  cos (x +  ?/)  =  cosx  cosy  —  sinx  sin  ?/. 

36.  cos(x  —  y)  =  cosx  cosy  +  sinx  siny. 
tan  X  +  tan  y 


37.  tan  (x  +  y)  = 

38.  tan(x  — ?/)  = 


1  —  tan  X  tan  y 

tan  X  —  tan  y 
1  +  tan  X  tan  y 


NEW  ANALYTIC  GEOMETRY 


39.  sill  2  a;  =  2  sin  a;  cosx;  cos2x  =  cos^x  —  sln^x  ;  taii2x  = 


2  tan  a; 
1  —  tan^  X 


40.  sin 


VI  —  COSX             X                   /l  +  ' 
;  cos  -  =  ±  -»  / 
2                2          \        I 


tan- 


xl 


—  cosx 


+  cosx 


2  \        2  2  \        2        '         2 

41.  sin-x  =  ^  —  ^cos2x  ;  cos^x  =  ^  +  ^cos2x. 

42.  sin  A  -  sin  i^  =  2  cos  J  (J.  +  B) sin  \(^A-B). 

43.  cos ^  —  cos I^  =  —  2 sin \{A-\-B) sin \{A  —  B). 

44.  Theorem.  Law  of  cosines.  In  any  triangle  the  square  of  a  side 
equals  the  sum  of  the  squares  of  the  two  other  sides  diminished  by  twice 
the  product  of  those  sides  by  the  cosine  of  their  included  angle  ;  that  is, 

a2  =  62  +  c2_  26c  cos  J.. 

46.  Theorem.   Area  of  a  triangle.   The  area  of  any  triangle  equals  one 
half  the  product  of  two  sides  by  the  sine  of  their  included  angle  ;  that  is, 
area  =  iab  sin  C  =  ^bc  sin  A  =  i  ca  sin  B. 

2.  Three-place  table  of  common  logarithms  of  numbers. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

000 

041 

079 

114 

146 

176 

204 

230 

255 

279 

2 

301 

322 

342 

362 

380 

398 

415 

431 

447 

462 

3 

477 

491 

505 

518 

532 

544 

556 

568 

580 

591 

4 

602 

613 

623 

634 

643 

653 

663 

672 

681 

690 

5 

699 

708 

716 

724 

732 

740 

748 

756 

763 

771 

6 

778 

785 

792 

799 

806 

813 

820 

826 

832 

839 

7 

845 

851 

857 

863 

869 

875 

881 

886 

892 

898 

8 

903 

908 

914 

919 

924 

929 

934 

939 

944 

949 

9 
10 

11 

954 

959 

964 

968 

973 

978 

982 

987 

991 

996 

000 

004 

009 

013 

017 

021 

025 

029 

033 

037 

041 

045 

049 

053 

057 

001 

064 

068 

072 

076 

12 

079 

083 

086 

090 

093 

097 

100 

104 

107 

111 

13 

114 

117 

121 

124 

127 

130 

134 

137 

140 

143 

14 

146 

149 

152 

155 

1.58 

161 

164 

167 

170 

173 

15 

176 

179 

182 

185 

188 

190 

193 

196 

199 

201 

16 

204 

207 

210 

212 

215 

218 

220 

223 

225 

228 

17 

230 

233 

236 

238 

241 

243 

246 

248 

250 

2.53 

18 

255 

258 

260 

262 

265 

267 

270 

272 

274 

276 

19 

279 

281 

283 

286 

288 

290 

292 

294 

297 

299 

FORMULAS  AND  TABLES  FOR  REFERENCE 
3.  Squares  and  cubes  ;  square  roots  and  cube  roots. 


No. 

Square 

Cube 

Square 
Root 

Cube 
Root 

No. 

Square 

Cube 

Square 
Root 

Cube 
Root 

1 

1 

1 

1.000 

1.000 

51 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.259 

52 

2,704 

140,608 

7.211 

3.732 

3 

9 

27 

1.732 

1.442 

53 

2,809 

148,877 

7.280 

3.756 

4 

16 

64 

2.000 

1.587 

54 

2,916 

157,464 

7.348 

3.779 

5 

25 

125 

2.236 

1.709 

55 

3,025 

166,375 

7.416 

3.802 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.825 

7 

49 

343 

2.645 

1.912 

57 

3,249 

185,193 

7.549 

3.848 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

7.615 

3.870 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

7.681 

3.892 

10 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

7.745 

3.914 

11 

121 

1,331 

3.316 

2.223 

61 

3,721 

226,981 

7.810 

3.9.36 

12 

144 

1,728 

3.464 

2!289 

62 

3,844 

238,328 

7.874 

3.957 

13 

169 

2,197 

3.605 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

196 

2,744 

3.741 

2.410 

64 

4,096 

262,144 

8.000 

4.000 

15 

225 

3,375 

3.872 

2.466 

65 

4,225 

274,625 

8.062 

4.020 

16 

256 

4,096 

4.000 

2.519 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.061 

18 

324 

5,832 

4.242 

2.620 

68 

4,624 

314,432 

8.246 

4.081 

19 

361 

6,859 

4.358 

2.668 

69 

4,761 

328,509 

8.306 

4.101 

20 

400 

8,000 

4.472 

2.714 

70 

4,900 

343,000 

8.366 

4.121 

21 

441 

9,261 

4.582 

2.758 

71 

5,041 

357,911 

8.426 

4.140 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.795 

2.843 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,824 

4.898 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

5.000 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

5,776 

438,976 

8.717 

4.235 

27 

729 

19,683 

5.196 

3.000 

77 

5,929 

456,533 

8.774 

4.254 

28 

784 

21,952 

5.291 

3.036 

78 

6,084 

474,552 

8.8.31 

4.272 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.290 

30 

900 

27,000 

5.477 

3.107 

80 

6,400 

512,000 

8.944 

4.308 

31 

961 

29,791 

5.567 

3.141 

81 

6,561 

531,441 

9.000 

4.326 

32 

1,024 

32,768 

5.656 

3.174 

82 

6,724 

551,368 

9.0.55 

4.344 

33 

1,089 

35,937 

5.744 

3.207 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

5.830 

3.239 

84 

7,056 

592,704 

9.165 

4.379 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.219 

4.396 

36 

1,296 

46,656 

6.000 

3.301 

86 

7,396 

636,056 

9.273 

4.414 

37 

l.:i69 

50,653 

6.082 

3.332 

87 

7,569 

658,503 

9..327 

4.431 

38 

1,444 

54,872 

6.164 

3.361 

88 

7,744 

681,472 

9.380 

4.447 

39 

1,521 

59,319 

6.244 

3.391 

89 

7,921 

704,969 

9.433 

4.464 

40 

1,600 

64,000 

6.324 

3.419 

90 

8,100 

729,000 

9.486 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.497 

42 

1,764 

74,088 

6.480 

3.476 

92 

8,464 

778,688 

9.591 

4.514 

43 

1,849 

79,507 

6.557 

3.503 

93 

8,649 

804,357 

9.643 

4.530 

44 

1,936 

85,184 

6.633 

3.530 

94 

8,836 

830,584 

9.695 

4.546 

45 

2,025 

91,125 

6.708 

3.556 

95 

9,025 

857,375 

9.746 

4.562 

46 

2,116 

97,336 

6.782 

3.583 

96 

9,216 

884,736 

9.797 

4.578 

47 

2,209 

103,823 

6.855 

3.608 

97 

9,409 

912,673 

9.848 

4..594 

48 

2,304 

110,592 

6.928 

3.634 

98 

9,604 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7.000 

.•5.6.59 

99 

9,801 

970,299 

9.949 

4.626 

50 

2,500 

125,000 

7.071 

3.()S4 

100 

10,000 

1,000,000 

10.000 

4.641 

NEW  ANALYTIC  GEOMETRY' 


4.   Natural  values  of  trigonometric  functions. 


Angle  in 
Radians 

Angle  in 
Degrees 

sin 

cos 

tan 

cot 

.000 
.017 
.035 
.052 
.070 
.087 
.174 
.262 
.349 
.436 
.524 
.611 
.698 
.785 

0° 

1° 

2° 

3° 

4° 

5° 

10° 

15° 

20° 

25° 

30° 

35° 

40° 

45° 

.000 
.017 
.035 
.0.52 
.070 
.087 
.174 
.259 
.342 
.423 
.500 
.574 
.643 
.707 

1.000 
.999 
.999 
.999 
.998 
.996 
.985 
.966 
.940 
.906 
.866 
.819 
.706 
.707 

.000 
.017 
.035 
.052 
.070 
.088 
.176 
.268 
.304 
.460 
.577 
.700 
.839 
1.000 

00 

57.29 

28.64 

19.08 

14.30 

11.43 

5.67 

3.73 

2.75 

2.14 

1.73 

1.43 

1.19 

1.00 

90° 
89° 
88° 
87° 
86° 
85° 
80° 
75° 
70° 
65° 
60° 
55° 
50° 
45° 

1.571 

1.553 

1.536 

1.518 

1.501 

1.484 

1.396 

1.309 

1.222 

1.134 

1.047 

.960 

.873 

.785 

cos 

sin 

cot 

tan 

Angle  in 
Degrees 

Angle  in 
Radians 

5.  Logarithms  of  trigonometric  functions. 


Angle  in 
Radians 

Angle  in 
Degrees 

log  sin 

log  cos 

log  tan 

log  cot 

.000 
.017 
.035 
.052 
.070 
.087 
.174 
.262 
.349 
.436 
.524 
.611 
.698 
.785 

0° 

1° 

2° 

3° 

4° 

5° 

10° 

15° 

20° 

25° 

30° 

35° 

40° 

46° 

8.242 
8.543 
8.719 
8.844 
8.940 
9.240 
9.413 
9.. 534 
9.020 
9.099 
9.759 
9.808 
9.850 

0.000 
9.999 
9.999 
9.999 
9.999 
9.998 
9.993 
9.985 
9.973 
9.957 
9.9.38 
9.913 
9.884 
9.850 

8.242 
8.543 
8.719 
8.845 
8.942 
9.246 
9.428 
9.561 
9.669 
9.701 
9.845 
9.924 
0.000 

1.758 
1.457 
1.281 
1.155 
1.058 
0.754 
0.572 
0.439 
0.331 
0.239 
0.105 
0.080 
0.000 

90° 
89° 
88° 
87° 
86° 
85° 
80° 
75° 
70° 
66° 
60° 
56° 
60° 
45° 

1.571 
1.653 
1.536 
1.518 
1.501 
1.484 
1.396 
1.309 
1.222 
1.134 
1.047 
0.960 
0.873 
0.785 

log  cos 

log  sin 

log  cot 

log  tan 

Angle  in 
Degrees 

Angle  in 
Radians 

FORMULAS  AND  TABLES  FOR  REFERENCE 
6.  Natural  values.    Special  angles. 


Angle  in 
Radians 

Angle  in 
Degrees 

sin 

cos 

tan 

cot 

sec 

esc 

0 

0° 

0 

1 

0 

00 

1 

CO 

iTT 

90° 

1 

0 

CO 

0 

00 

1 

TT 

180° 

0 

-1 

0 

CO 

- 1 

CO 

|7r 

270° 

-1 

0 

CO 

0 

CO 

-1 

27r 

360° 

0 

1 

0 

oo 

1 

CO 

Angle  in 
Radians 

Angle  in 
Degrees 

sin 

cos 

tan 

cot 

sec 

CSC 

0 

0° 

0 

1 

0 

00 

1 

CO 

i-^ 

30° 

i 

iVs 

iVs 

V3 

sV3 

2 

i^ 

45° 

*V2 

iV2 

1 

1 

V"2 

V^ 

l^ 

G0° 

i  V3 

i 

v^ 

^V3 

2 

§  V3 

i-n- 

90° 

1 

0 

CO 

0 

CO 

1 

7.  Rules  for  signs  of  the  trigonometric  functions. 


Quadrant 

sin 

cos 

tan 

cot 

sec 

CSC 

First  .... 

+ 

+ 

+ 

+ 

+ 

+ 

Second    .     .     . 

+ 

- 

- 

- 

- 

+ 

Thinl       .     .     . 

- 

- 

+ 

+ 

- 

- 

Finirth     . 

- 

+ 

- 

- 

+ 

- 

CHAPTER  II 

CARTESIAN  COORDINATES 

8.  Directed  line.  Let  A''A'  be  an  indefinite  straight  line, 
and  let  a  point  O,  which  we  shall  call  the  origin,  be  chosen 
upon  it.  Let  a  unit  of  length  be  adopted,  and  assume  that 
lengths  measured  frou)  O  to  the  right  are  positive,  and  to 
the  left  negative. 

Then  any  real  number,  if  taken  as  the  measure  of  the 
length  of  a  line  OP,  will  determine  a  point  P  on  the  line.  Con- 
versely, to  each  point  _5_4_3_2_]  0+1+2+3+4+5  unu 
P  on  the  line  will  cor-  y,  — ^^ — ''~'^~o~^ — *_o— ^— — ^  <« 
respond  a  real  num- 
ber, namely  the  measure  of  the  length  OP,  witli  a  positive 
or  negative  sign  according  as  P  is  to  tlie  right  or  left  of 
the  origin. 

The  direction  established  upon  X'X  by  passing  from  the 
origin  to  the  points  corresponding  to  the  positive  numbers  is 

called  the  positive  direction   — ^ = , = *-*. 

on  the  line.   A  dii'ected  line 

is  a  straight  line  upon  which  an  origin,  a  unit  of  length,  and  a 

positive  direction  have  been  assumed. 

An  arrowhead  is  usually  placed  upon  a  directed  line  to  indi- 
cate the  positive  direction. 

If  A  and  B  are  any  two  points  of  a  directed  line  such  that 

OA  =a,      OB  =  h, 

then  the  length  of  the  segment  AB  is  always  given  hy  h  —  a\ 
that  is,  the  length  of  AB  is  the  difference   of  tlie  numbers 


CARTESIAN   COORDINATES  9 

corresponding  to  B  and  .4.    This  statement  is  evidently  equiv- 
alent to  the  following  definition  : 

For  all  positions  of  two  points  A  and  B  on  a  dii^ected  line,  the 
length  AB  is  given  by 
(1)  AB  =  OB  -  OA, 

where  O  is  the  origin. 

(II)  (HI)  (IV) 

0+3  -3       0  +5       -6       -2     0 


BOA         A     0  B      B       A   O     . 

The  above  definition  is  illustrated  in  each  of  the  four  figures, 
as  follows : 

I.  AB=OB-OA  =  Q-Z=-\-^;  BA-OA-OB  =  ^-Q  =  -Z  ; 
II.  AB=OB-OA  =  -^-Z=-l  ;  BA  =  OA-OB  =  ^-{-A)  =  +  l ; 

III.  AB=OB-OA  =  -\-b-{-Z)  =  -\-?>;  BA  =  0A-0B=-S-rj  =  -8  ■ 

IV.  AB=OB-OA  =  -6-{-2)  =  -'i;BA  =  OA-OB=-2-{-6)  =  +4. 

The  following  properties  of  lengths  on  a  directed  line  are 
obvious : 

(2)  AB  =  ~BA. 

(3)  AB  is  positive  if  the  direction  from  A  to  B  agrees  with 
the  positive  direction  on  the  line,  and  negative  if  in  the  con- 
trary direction. 

The  phrase  "distance  between  two  points"  should  not  be  used  if 
these  points  lie  upon  a  directed  line.  Instead,  we  speak  of  the  length 
AB,  remembering  that  the  lengths  ABaud  BA  are  not  equal,  but  that 
AB=-BA. 

9.  Cartesian*  coordinates.  Let  .Y'A' and  F'y  be  two  directed 
lines  intersecting  at  O,  and  let  P  be  any  point  in  their  plane. 
Draw  lines  through  P  parallel  to  X'X  and  Y'Y  respectively. 
Then,  if  OAf=a,      ON  =  b, 

*  So  called  after  Rene  Descartes,  1596-16.50,  who  first  introduced  the  idea 
of  coordinates  into  the  study  of  geometry. 


10 


NEW  ANALYTIC  GEOMETRY 


the  numbers  a,  b  are  called  the  Cartesian  coordinates  of  P,  a  the 

abscissa  and  h  the  ordinate.    The  directed  lines  X'X  and  y'Fare 

called  the  axes  of  coordi- 

nates,  X'X  the   axis   of  / 

/  P 

abscissas,    Y'Y  the  axis 

of  ordinates,  and  their  in- 
tersection O  the  origin. 

The  coordinates  a,  b  of 
P  are  written  {q,,  b),  and 
the  symbol  P(a,  b)  is  to 
be  read,  "  The  point  P, 
whose  coordinates  are  a 
and  b." 

Any  point  P  in  the 
l)lane  determines  two  numbers,  the  coordinates  of  P.  Con- 
versely, given  two  real  numbers  a'  and  b',  then  a  point  P'  in 
the  plane  may  always  be  constructed  whose  coordinates  are 
(a',  b').  Por  lay  off  OM'  =  a',  ON'  =  b',  and  draw  lines  parallel 
to  the  axes  through  M'  and  N'.  These  lines  intersect  at  P' 
(a',  Z»').    Hence 

Every  point  determines  a  pair  of  real  numbers,  and,  conversely, 
a  pair  of  real  numbers  determines  a  2^0 int. 

The  imaginary  numbers  of  algebra  have  no  place  in  this 
representation,  and  for  this  reason  elementary  analytic  geome- 
try is  concerned  only  with  the  real  numbers  of  algebra. 

10.  Rectangular  coordinates.  A  rectangular  system  of  coordi- 
nates is  determined  when  the  axes  A''A  and  l''Fare  perpendicular 
to  each  other.  This  is  the  usual  case,  and  will  be  assumed  unless 
otherwise  stated. 

The  work  of  plotting  points  in  a  rectangular  system  is  much 
simplified  by  the  use  of  coordinate  ox  plotting  pi'^^P'i^'i  constructed 
by  ruling  off  the  plane  into  equal  scpiares,  the  sides  being 
parallel  to  the  axes. 


CARTESIAN  COORDINATES 


11 


In  the  figure  several  points  are  plotted,  the  unit  of  length 
being  assumed  equal  to  one  division  on  each  axis.  The  method 
is  simply  this : 

Count  off  from  O  along  A' 'A'  a  number  of  divisions  equal  to 
the  given  abscissa,  and  then  from  the  point  so  determined  a 


A"' 


f-9,- 


Kie) 


(0, 


(6, 


(10 


0) 


X 


Y\ 


number  of  divisions  up  or  down  equal  to  the  given  ordinate, 
observing  the 

Rule  for  signs : 

Abscissas  are  positive  or  negative  according  as  tlietj  are  laid 
off  to  the  right  or  left  of  the  origin.  Ordinates  are  positive  or 
negative  according  as  they  are  laid  off  above 
or  below  the  axis  of  x. 

Rectangular  axes  divide  the  plane  into 
four  portions  called  quadrants ;  these  are 
numbered  as  in  the  figure,  in  which  the 
proper  signs  of  the  coordinates  are  also 
indicated. 

As    distinguished   from    rectangular    coordinates,    the    term 
oblique    coordinates    is    employed    when    the    axes     are     not 


Second 


X'  0 

Third 


First 
(f.+) 


A' 


Fourth 

(f,-) 


12  NEW  ANALYTIC  GEOMETRY 

perpendicular,  as  in  the  figure  of  Art.  9.  The  rule  of  signs 
given  above  applies  to  this  case  also.  Note,  however,  in 
plotting,  that  the  ordinate  MP  is  drawn  iKirallel  to  OY. 

In  the  following  problems  assume  rectangular  coordinates 
unless  the  contrary  is  stated. 

PROBLEMS 

1.  Plot  accurately  the  points  (3,  2),  (3,  -  2),  (-  4,  3),  (6,  0),  (-  5,  0), 
(0,  4). 

2.  What  are  the  coordinates  of  the  origin  ?  J.jis.  (0,  0). 

3.  In  what  quadrants  do  the  following  points  lie  if  a  and  h  are  posi- 
tive numbers  :  (-  a,  6)  ?  (-  a,  -  6)  ?  (6,  -  a)  ?  (o,  h)  ? 

4.  To  what  quadrants  is  a  point  limited  if  its  abscissa  is  positive  ? 
negative  ?   if  its  ordinate  is  positive  ?   negative  ? 

5.  Draw  the  triangle  whose  vertices  are  (2,  —  1),  (—  2,  5),  (—  8,-4). 

6.  Plot  the  points  whose  oblique  coordinates  are  as  follows,  when  the 
angle  between  the  axes  is  60°:  (2,  -  3),  (3,  -  2),  (4,  5),  (-  6,  -7),  (-  8,  0), 
(9,  -5),  (-6,  2). 

7.  Draw  the  quadrilateral  whose  vertices  are  (0,  —  2),  (4,  2),  (0,  6), 
(—  4,  2),  when  the  angle  between  the  axes  is  60°. 

8.  If  a  point  moves  parallel  to  the  axis  of  x,  which  of  its  coordinates 
i-emains  constant  ?    If  parallel  to  the  axis  of  ?/  ? 

9.  Can  a  point  move  when  its  abscissa  is  zero  ?  Where  ?  Can  it  move 
when  its  ordinate  is  zero  ?  Where  ?  Can  it  move  if  both  abscissa  and 
ordinate  are  zero  ?    Where  will  it  be  ? 

10.  Where  may  a  point  be  found  if  its  abscissa  is  2  ?  if  its  ordinate  is 
-3? 

11.  Where  do  all  those  points  lie  whose  abscissas  and  ordinates  are 
equal  ? 

12.  Two  sides  of  a  rectangle  of  lengths  a  and  6  coincide  with  the  axes 
of  X  and  y  respectively.  What  are  the  coordinates  of  the  vertices  of  the 
rectangle  if  it  lies  in  the  first  quadrant  ?  in  the  second  quadrant  ?  in  the 
third  quadrant  ?  in  the  fourth  quadrant  ? 

13.  Construct  the  quadrilateral  whose  vertices  are  (—  3,  6),  (—  3,  0), 
(3,  0),  (3,  6).  What  kind  of  a  quadrilateral  is  it  ?  What  kind  of  a  quad- 
rilateral is  it  when  the  axes  are  oblique  ? 


CARTESIAN  COORDINATES 


13 


14.  Show  that  {x,  y)  and  (x,  —  y)  are  symmetrical  with  respect  to  X'X  ; 
(x,  y)  and  (—  x,  y)  with  respect  to  Y'Y;  and  (x,  y)  and  (—  x,  —  y)  with 
respect  to  the  origin. 

15.  A  line  joining  two  points  is  bisected  at  the  origin.  If  the  coordinates 
of  one  end  are  (a,  —  6),  what  will  be  the  coordinates  of  the  other  end  ? 

16.  Consider  the  bisectors  of  the  angles  between  the  coordinate  axes. 
What  is  the  relation  between  the  abscissa  and  ordinate  of  any  point  of  the 
bisector  in  the  first  and  third  quadrants  ?  second  and  fourth  quadrants  ? 

17.  A  square  whose  side  is  2  a  has  its  center  at  the  origin.  "What  will 
be  the  coordinates  of  its  vertices  if  the  sides  are  parallel  to  the  axes  ?  if 
the  diagonals  coincide  with  the  axes  ? 

Ans.  {a,  a),  (a,  —a),  (—  a,  —  a),  (—  a,  a)  ; 

(aV2,  O),  (-aV2,  O),  (O,  aV2),  (O,  -  aV2). 

18.  An  equilateral  triangle  whose  side  is  a  has  its  base  on  the  axis  of 
X  and  the  opposite  vertex  above  X'X.  What  are  the  vertices  of  the  tri- 
angle if  the  center  of  the  base  is  at  the  origin  ?  if  the  lower  left-hand 
vertex  is  at  the  origin  ? 

(0,  0),  (a,  0),  (-, 


Ans. 


11.  Lengths. 


|,o),(-^,o),(o, 


/a   aV3\ 
\2'"T~/ 


Then  in  the  figure  OM,  =  x 


Consider  any  two  given  points 


2V^2'  ^a)" 


We  may  now  easily  prove  the  important 
Theorem.   The  length  I  of  the  line 

Joining  two  points  P.^(x^,  i/^),  P.^i^^'  V-i) 

is  given  by  the  formula  ' 


OM=x.^,M^P^=y^,M^P^=y.^ 


ro(X2,;?/„) 


(I)     l=^{x^-x,y  +  {y^-y,)\ 
Proof.    Draw  lines  through  P^  and 

P^  parallel  to  the  axes  to  form  the 

right  triangle  P^SP^. 
Then 


P^S  =  OM,  - 
SP  =  M  P 

2  2      2 


P^P 


=V^' 


+  i\s- 


and  hence 


^  =  V(.r^-a-/+(/A-.y,y^ 


Q.E.  D. 


14 


NEW  ANALYTIC   GEOMETRY 


The  same  method  is  used  in  deriving  (I)  for  any  positions  of 
P  and  P, ;  namely,  we  construct  a  right  triangle  by  drawing 
lines  parallel  to  the  axes  through  P^  and  P^.  The  horizontal 
side  of  this  triangle  is  equal  to  the  difference  of  the  abscissas 
of  P  and  Pg,  while  the  vertical  side  is  equal  to  the  difference  of 
the  ordinates.  The  required  length  is  then  the  square  root  of  the 
sum  of  the  squares  of  these  sides,  which  gives  (I).  A  number 
of  different  figures  should  be  drawn  to  make  the  method  clear. 

EXAMPLE 

Find  the  length  of  the  line  joining  the  points  (1,  3)  and  (—  5,  5). 

Solution.   Call  (1,  3)  P^,  and  (-5,  5)  P„. 

Then  x^  =  1,     2/i  =  3, 

and  Xo  =  —  5,     y„  =  b; 

and  substituting  in  (I),  we  have 


Y' 

(-5 

5-) 

r^^ 

■•^ 

(1 

3-) 

.Y' 

0 

X 

Y 

I  =  V(l  +  5)2  +  (3  -  5)2  =  V40=  2  VlO. 

It  should  be  noticed  that  we  are  simply 
finding  the  hypotenuse  of  a  right  triangle 
whose  sides  are  6  and  2. 

Remark.  The  fact  that  formula 
(I)  is  true  for  all  positions  of  the  points  P^  and  7^„  is  of  funda- 
mental importance.  The  application  of  this  formula  to  any 
given  problem  is  therefore  simply  a  matter  of  direct  substitu- 
tion. In  deriving  such  general  formulas  it  is  most  convenient 
to  draw  the  figure  so  that  the  points  lie  in  the  first  quadrant, 
or,  in  general,  so  that  all  the  quantities  assumed  as  knoirn  shall 
he  positive. 


PROBLEMS 

1.  Find  the  lengths  of  the  lines  joining  the  following  points : 
(a)  (-  4,  -  4)  and  (1,  3). 


Ans.  Vli. 
Aivi.  VlO. 


(b)  (-V2,  V3)  and  (Vs,  V2). 

(c)(0,0)and(|.^^). 

(il)   {(t  +  b,  c  +  a)  and  (c  +  a,  b  +  c).  Ans.   V(b  —  c)2  +  (a  —  6)2. 


Ans.  a. 


CARTESIAN  COORDINATES  15 

2.  Find  the  lengths  of  the  sides  of  the  following  triangles  : 

(a)  (0,  6),  (1,  2),  (3,  -  5). 

(b)  (1,0),  (-1,-5),  (-1,  -8). 

(c)  («,  6),  (6,  c),  (c,  d). 

(d)  (a,  -  6),  (6,  -  c),  (c,  -  d). 

(e)  (0,2/),  (-X,  _y),  (-x,  0). 

3.  Find  the  lengths  of  the  sides  of  the  triangle  whose  vertices  are 
(4,  3),  (2,  -  2),  (-  3,  5). 

4.  Show  that  the  points  (1,  4),  (4,  1),  (6,  5)  are  the  vertices  of  an 
isosceles  triangle. 

5.  Show  that  the  points  (2,  2),  (-  2,  -  2),  (2  Vs,  -  2  Vs)  are  the 
vertices  of  an  equilateral  triangle. 

6.  Show  that  (3,  0),  (6,  4),  (—  1,  3)  are  the  vertices  of  a  right  triangle. 
What  is  its  area  ? 

7.  Prove  that  (-  4,  -  2),  (2,  0),  (8,  6),  (2,  4)  are  the  vertices  of  a 
parallelogram.    Also  find  the  lengths  of  the  diagonals. 

8.  Show  that  (11,  2),  (6,  -  10),  (-  6,  -  5),  (-  1,  7)  are  the  vertices 
of  a  square.   Find  its  area. 

9.  Show  that  the  points  (1,  3),  (2,  V6),  (2,  —Vo)  are  equidistant 
from  the  origin  ;  that  is,  show  that  they  lie  on  a  circle  with  its  center  at 
the  origin  and  its  radius  equal  to  the  VlO. 

10.  Show  that  the  diagonals  of  any  rectangle  are  equal. 

11.  Find  the  perimeter  of  the  triangle  whose  vertices  are  (a,  6),  (—  a,  6), 
(-«,-&)•  '        . 

12.  Find  the  perimeter  of  the  polygon  formed  by  joining  the  following 
points  two  by  two  in  order :  (6,  4),  (4,  -  3),  (0,  —  1),  (—  5,  —  4),  (—  2,  1). 

13.  One  end  of  a  line  whose  length  is  13  is  the  point  (—  4,  8) ;  the 
ordinate  of  the  other  end  is  3.    What  is  its  abscissa  ?        Ans.  8  or  —  16. 

14.  What  equation  must  the  coordinates  of  the  point  (x,  y)  satisfy  if 
its  distance  from  the  point  (7,  —  2)  is  equal  to  11  ? 

15.  What  equation  expresses  algebraically  the  fact  that  the  point  (x,  y) 
is  equidistant  from  the  points  (2,  3)  and  (4,  5)  ? 

16.  Find  the  length  of  the  line  joining  Pi(x,,  ?/,)  and  P^i'f-zi  V-z)  when 
the  coordinates  are  oblique. 

Hint.    Use  the  law  of  cosines,  44,  p.  4. 


16 


NEW  ANALYTIC  GEOMETRY 


''  12.  Inclination  and  slope.  The  angle  between  two  intersecting 
directed  lines  is  defined  to  be  the  angle  made  by  their  positive 
directions.  In  the  figures  the  angle  between  the  directed  lines 
is  the  angle  marked  6. 

If  the  directed  lines  are  parallel,  then  the 
angle  between  them  is  zero  or  180°,  according 
as  the  positive  directions  agree  or  do  not 
agfree. 

Evidently  the  angle  between  two  directed 
lines  may   have  any  value    from  0  to  180° 
inclusive.     Reversing  the  direction  of  either 
directed  line  changes  B  to  the  supplement  180° 
directions  are  reversed,  the  angle  is  unchanged. 

When  it  is  desired  to  assign  a  positive  direction  to  a  line 
intersecting  X'X,  we  shall  always  assume  the  upward  direction 
as  positive.  * 

The  inclination  of  a  line  is  the  angle  be- 
tween the  axis  of  x  and  the  line  when  the 
latter  is  given  the  upward  direction. 


e.    If  both 


e^Q 


This  amounts  to  saying  that  the  inclination  is  the 
angle  above  the  x-axis  and  to  the  right  of  the  given  line,  as  in  the 


The  slojje  of  a  line  is  the  tangent  of  its  inclination. 

The  inclination  of  a  line  will 
be  denoted  by  the  Greek  letter  a, 
ftj,  «:.„  a',  etc.  ("alpha,"  etc.);  its 
slope  by  m,  m^,  m.^,  vi',  etc.,  so 
that  VI  =  tan  a,  iti^  =  tan  a^,  etc. 

The  inclination  may  be  any 
angle  from  0  to  180°  inclusive. 

The  slope  may  be  any  real  num- 
ber, since  the  tangent  of  an  angle 
in  the  first  two  qiiadrants  may  be  any  number  positive  or  nega- 
tive.   The  slope  of  a  line  parallel  to  A' 'A'  is  of  course  zero,  since 


CARTESIAN  COORDINATES 


17 


the  inclination  is  0  or  180°.    For  a  line  parallel  to  Y'Y  the  slope 
is  infinite. 

Theorem.    The  slojie  m  of  the  line  passing  through  two  points 
i\{^\^  2/i)'  ^2(*'2'  2/2)  '^  9^^^'^  h 

(II)  '"=^- 

-*1  •*2 

Proof.    In  the  figure 

Draw  P^S  parallel  to  OX.    Then  in  the  right  triangle  P.2''^P^, 
since  angle  P,P„S  =  a,  we  have 

12  7  S'  y- 1 

SP, 

(1) 

But 


and 


971  =  tan  a  =  — — ;  • 

2 

y(px 

=  MJ-^-M^P=y^-y.^; 

—4s 
1 

P^S  =  M^M^ 

0 

/^        M2 

Mix 

=  OM,  —  OM,  =  X,  —  .7- . 

X 

Substituting  these  values  in  (1)  gives  (II).  Q.  e.  d. 

The  student  should  derive  (II)  when  a  is  obtuse.* 
We  next  derive  the  conditions  for  parallel  lines  and  for  per- 
pendicular lines  in  terms  of  their  slopes. 

Theorem.  If  two  lines  are  parallel,  their  slopes  are  equal ;  if 
perpendicular,  the  slope  of  one  is  the  negative  reciprocal  of  tht 
slope  of  the  other,  and  conversely. 

Proof.  Let  a^  and  a^  be  the  inclinations  and  m^  and  in  the 
slopes  of  the  lines. 

If  the  lines  are  parallel,  a^  =  a,^.    .'.  vi^  =  m,^. 

*  To  construct  a  line  passing  through  a  given  point  P^  whose  slope  is  a  pos- 
itive fraction  - ,  we  mark  a  point  S  h  units  to  the  right  of  I\  and  a  point  P^ 

a  units  above  iS,  and  draw  P^P^-   If  the  slope  is  a  negative  fraction,  —  ,  then 
plot  S  6  units  to  the  left  of  P^.  ^ 


18 


NEW  ANALYTIC  GEOMETRY 


If  the  lines  are  perpendicular,  as  in  the  figure, 

TT 

«,=  77  +  «r 
.  m.^  =  tan  «,  =  tan  (  —  +  ^j ) 

=  -  cot  a^       (by  31,  p.  3) 
(By  26,  p.  3) 


tan  a^ 

J_ 

m. 


The  converse  is    proved   by  retracing  the   steps  with  the 
assumption,  in  the  second  part,  that  a,  is  greater  than  a^. 


PROBLEMS 

1.  Find  the  slope  of  the  line  joining  (1,  3)  and  (2,  7).  Ans.    4. 

2.  Find  the  slope  of  the  line  joining  (2,  7)  and  (—  4,  —  4).    Aixs.    y. 

3.  Find  the  slope  of  the  line  joining  (V3,  V^)  and  (— V2,  Vs). 

Ans.   2V6-  5. 

4.  Find  the  slope  of  the  line  joining  (a  +  6,  c  +  «),  (c  +  a,  6  +  c). 

.         b  —  a 

Ans. 

c-  b 

5.  Find  tlie  slopes  of  the  sides  of  the  triangle  whose  vertices  are  (1,  1), 
(-1,  -1),  (V3,  -V3).  ^^^     ^    I+V3    I-V3 

'i-Vs'i  +  Vs 

'    6.  Prove  by  means  of  slopes  that  (—4,  —  2),  (2,  0),  (8,  6),  (2,  4)  are 
the  vertices  of  a  parallelogram. 

■  7.  Prove  by  means  of  slopes  that  (3,  0),  (6,  4),  (—  1,  3)  are  the  vertices 
of  a  right  triapgle. 

'  8.  Prove  by  means  of  slopes  that  (0,  -  2),  (4,  2),  (0,  6),  (-  4,  2)  are 
the  vertices  of  a  rectangle,  and  hence,  by  (I),  of  a  square. 

9.  Prove  by  means  of  their  slopes  that  the  diagonals  of  the  square  in 
Problem  8  are  perpendicular. 

V  10.  Prove  by  means  of  slopes  that  (10,  0),  (5,  5),  (5,  —  5),  (—  5,  5)  are 
the  vertices  of  a  trapezoid. 

11.  Show  that  the  line  joining  (a,  b)  and  (c,  —  d)  is  parallel  to  the  line 
joining  (—  a,  —  b)  and  (—  c,  d). 


CARTESIAN  COORDINATES  19 

12.  Show  that  the  line  joining  the  origin  to  (a,  b)  is  perpendicular  to 
the  line  joining  the  origin  to  (—  b,  a). 

"   13.  What  is  the  inclination  of  a  line  parallel  to  Y'Y?  perpendicular 
to  FT? 
'    14.  What  is  the  slope  of  a  line  parallel  to  Y'Y  ?  perpendicular  to  Y'Y? 

15.  What  is  the  inclination  of  the  line  joining  (2,  2)  and  (—  2,  —  2)  ? 

Ans.    — . 
4 

16.  What  is  the  inclination  of  the  line  joining  (—  2,  0)  and  (—  5,  3)  ? 

Stt 

Ans. 

4 

17.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (4,  y/S)  ? 

Ans.    -. 

18.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (2,  -\/3)  ? 

Ans. 

3 

19.  Whatistheinclinationof  the  line  joining  (0,  —  4)  and(—  a/3,—  -5)  ? 

Ans.    -■ 
6 

20.  What  is  the  inclination  of  the  line  joining  (0,  0)  and  (—  -y/s,  1)  ? 

Ans. 

U 
^    21.  Prove  by  means  of  slopes  that  (2,  3),  (1,  —  3),  (3,  9)  lie  on  the 
same  straight  line. 

22.  Prove  that  the  points  (a,  b  +  c),  (5,  c  +  a),  and  (e,  a  +  b)  lie  on  the 
same  straight  line. 

23.  Prove  that  (1,  .5)  is  on  the  line  joining  the  points  (0,  2)  and  (2,  8) 
and  is  equidistant  from  them. 

■   24.  Prove  that  the  line  joining  (3,  —  2)  and  (.5,  1)  is  perpendicular  to 
the  line  joining  (10,  0)  and  (13,  -  2). 

13.  Point  of  division.    Let  P^  and  P,  be  two  fixed  points  on  a 
directed  line.    Any  third  point  on  the  line,  as  P  or  P',  is  said 

"  to  divide  the  line  into     — _c . ^ o 

two    segments,      and     is  ' 

called  a  point  of  division.  The  division  is  called  internal  or 
external  according  as  the  point  falls  within  or  without  PJ^.^ 
The  position  of  the  point  of  division  depends  upon  the  nitio 


20  NEW  ANALYTIC   GEOMETRY 

of  its  distances  from  P^  and  P.^.  Since,  however,  the  line  is 
directed,  some  convention  must  be  made  as  to  the  manner  of 
reading  these  distances.  We  therefore  adopt  the  rule  : 

If  P  is  a  point  of  division  on  a  directed  line  passing  through 

P^  and  P^,  then  P  is  said  to  divide  P^P.2  into  the  segments  P^P 

p  p 

and  PP„.   The  ratio  of  division  is  the  value  of  the  ratio  *  — ^  • 

PP^ 

We  shall  denote  this  ratio  by  X  (G-reek  letter  "lambda"), 
that  is,  p  p 

If  the  division  is  internal,  PJ^  and  PP^  agree  in  direction 
and  therefore  in  sign,  and  X  is  therefore  positive.  In  external 
division  X  is  negative. 

„,  .  p  N    J.1  o  -1<X<0  A=0        A>0         X=.o      _x<X<-l 

The  Sign  or  X  thereiore = 

PP. 
indicates    whether    the  '  " 

point  of  division  P  is  within  or  without  the  segment  P^P.,; 

and  the  numerical  value  determines  whether  P  lies  nearer  P^ 

or  P.^.    Tlie  distribution  of  X  is  indicated  in  the  figure.  ' 

That  is,  X  may  have  any  positive  value  between  P^  and  P^, 
any  negative  value  between  0  and  —  1  to  the  left  of  P^,  and  any 
negative  value  between  —  1  and  —  go  to  the  right  of  P.,.  The 
value  —  1  for  X  is  excluded. 

Introducing  coordinates,  we  next  prove  the 

Theorem.  Point  of  division.  The  coordinates  (x,  y)  of  the  point 
of  division  P  on  the  line  joining  P^(x^,  y^,  P.^^^.^,  y.^,  such  that 
the  ratio  of  the  segments  Is 

are  given  by  the  formulas 

(III)      r   ^^  +  ^^^     ,.    y.  +  ^y^ 


Y 

y 

^ 

J 

t  h  > 

/  0 

3/i 

M  31,  X 

1+A  1  +  A 

«  To  assist  the  memory  in  writing  down  this  ratio,  notice  that  the  jtoint  of 
division  /'  is  written  last  in  tlie  numerator  and, first  in  tlie  denominator. 


CARTESIAN  COORDINATES  21 

P,P 


Proof.    Given  X  = 

PP 

2 

Draw  the  ordinates  M^P^,  MP,  and  M^P,^.  Then,  by  geometry, 
these  ordinates  will  intercept  proportional  segments  on  the 
transversals  P^P,^  and  OX ;  that  is,* 

(1)  M,M_P,P 

But 

and,  by  hypothesis, 

Substituting  in  (1),  — 

Clearing  of  fractions  and  solving  for  x, 

X.  +  Xa^'o 
X  = 


MM^       PP,^ 

M^M  =0M  - 

OM^  =  X  —  x^, 

MM.^  =  OM.^  - 

-  OM  =  x.^  —  X, 

PP,  ~    ' 

x-x,  _^ 

1  +  A 


Similarly,  by  drawing  the  abscissas  of  P.^,  P,  and  P.,  to  the 

axis  or  y  we  may  prove       ?/  =  — ~-  Q.e.d. 

1+A 

Corollary.  Middle  point.  T/ie  coordinates  (x,  y)  of  the  middle 
point  of  tlte  line  Joining  Pj(a"j,  ^j),  Po(:''o,  y^  are  found  hy  taking 
the  averages  of  the  given  abscissas  and  ordinates;  that  is, 

(IV)  x=l(x^-\-x^),     y  =  l(y,  +  y^). 

P  P 

For  if  P  is  the  middle  point  of  P^P.,,  then  \  =  -^—  =  1,  and 

substituting  A  =  1  in  (III)  gives  (IV).  '^ 

To  apply  (III),  mark  the  point  of  division  P,  the  extremities 
of  the  line  to  be  divided  P^  and  P„  and  make  sure  that  the  value 
of  X  satisfies  X  =  P^P  h-  PP,. 

*  Care  must  be  taken  to  read  the  segments  on  the  transversals  (since  we  are 
dealing  with  directed  lines)  so  that  they  all  have  positive  directions 


22 


NEW  ANALYTIC    GEOMETRY 


EXAMPLES 
1.  Find  tlie  point  P  dividing  P^(— 1,   - 

Solution.    By  the  statement, 

PjP  _      1 
PP^-      4 

Hence,  applying  (III),  x■^^=—l,y^=—  (i, 
X2  =  3,  2/2  =  0. 
-1-1.3       -|_ 


6),   P.,  (3.  0)   in  tlie  ratio 

n 


-6-^.0 

3 

4 

1-i 

f 

I  — 2i 


y  = 


Hence  P  is  (-  2i,  -  8).  ^tw. 

The  result  is  checked  by  plotting.    The  point  P  lies  outside  the  seg- 
ment P^P.,  and  the  length  of  PP2  is  four  times  that  of  PPi- 

2.  Center  of  gravity  of  a  triangle.    Find  the  coordinates  of  the  point 
of  intersection  of  the  medians  of  a  triangle  whose 
vertices  are  (x^,  y{),  {x^,  y^),  (Xg,  y^). 

Solution.  By  plane  geometry  we  have  to  find 
the  point  P  on  the  median  AD  such  that  AP  — 
f  AD  ;  that  is,  AP  :  PZ) : :  2  : 1,  or  \  =  2. 

By  the  corollary,  D  is  [i(x2  +  X3),  |  (2/3  +  2/3)]. 

To  find  P,  apply  (III),  remembering  that  A 
corresponds  to  (x^,  y-^)  and  D  to  (x„  y„). 


J  (^i.2/1) 


(x2,yo) 


This  gives  x 


^Xi  +  2-^(X2  +  X,) 

1  +  2 

=  H^l  +  X2  +  X^), 


_yi  +  -^■i{y.2  +  Vs) 
1  +  2 

=  3(2/1  +  yo  +  Vs)-  ^ns. 


Hence  the  abscissa  of  the  intersection  of  the  medians  of  a  triangle  is 
the  average  of  the  abscissas  of  the  vertices,  and  similarly  for  the  ordinate. 

The  symmetry  of  these  answers  is  evidence  that  the  particular  median 
chosen  is  Immaterial,  and  the  formulas  therefore  prove  the  fact  of  the 
intersection  of  the  medians. 

The  result  just  found  admits  of  extension  to  any  polygon,  and,  with 
formulas  (IV),  illustrates  the  fact  that  the  cooi'dinates  of  centers  of 
gravity  are  found  by  taking  average  values. 


CARTESIAN  COORDINATES  23 

PROBLEMS 

1.  Find  the  coordinates  of  the  middle  point  of  the  line  joining  (4,  —  6) 
and  (-  2,  -  4).  Ans.  (1,  -  5). 

2.  Find  the  coordinates  of  the  middle  point  of  the  line  joining 
(a  +  b,  c  +  d)  and  (a  —  b,  d  —  c).  Ans.  {a,  d). 

3.  Find  the  middle  points  of  the  sides  of  the  triangle  whose  vertices 
are  (2,  3),  (4,  —  5),  and  (—  3,  —  0).    Also  find  the  lengths  of  the  medians. 

4.  Find  the  coordinates  of  the  point  which  divides  the  line  joining 
(—  1,  4)  and  (—  5,  —  8)  in  the  ratio  1 : 3.  Ans.  (-  2,  1). 

5.  Find  the  coordinates  of  the  point  which  divides  the  line  joining 
(_  3,  _  5)  and  (6,  9)  in  the  ratio  2  : 5.  Ans.  (-  ?,  —  1). 

6.  Find  the  coordinates  of  the  point  which  divides  the  line  joining 
(2,  6)  and  (-p  4,  8)  into  segments  whose  ratio  is  —  |.       Ans.  (—  22,  14). 

7.  Find  the  coordinates  of  the  point  which  divides  the  line  joining 
(—  3,  —  4)  and  (5,  2)  into  segments  whose  ratio  is  —  |.  Ans.  (—  19,  —  16). 

8.  Find  the  coordinates  of  the  points  which  trisect  the  line  joining 
the  points  (—  2,  —  1)  and  (3,  2).  Ans.  (—  |,  0),  (J,  1). 

9.  Prove  that  the  middle  point  of  the  hypotennse  of  a  right  triangle  i 
is  equidistant  from  the  three  vertices. 

10.  Show  that  the  diagonals  of  the  parallelogram  whose  vertices  are 
(1,  2),  (—  5,  —  3),  (7,  —  6),  (1,  —  11)  bisect  each  other.     . 

11.  Prove  that  the  diagonals  of  any  parallelogram  bisect  each  other. 

12.  Show  that  the  lines  joining  the  middle  points  of  the  opposite  sides 
of  the  quadrilateral  whose  vertices  are  (6,  8),  (—  4,  0),  (—  2,  —  6),  (4,—  4) 
bisect  each  other. 

13.  In  the  quadrilateral  of  Problem  12  show  by  means  of  slopes  that  the 
lines  joining  the  middle  points  of  the  adjacent  sides  form  a  parallelogram. 

14.  Show  that  in  the  trapezoid  whose  vertices  are  (—  8,  0),  {—  4,  —  4), 
(—  4,  4),  and  (4,  —  4)  the  length  of  the  line  joining  the  middle  points  of 
the  nonparallel  sides  is  equal  to  one  half  the  .sum  of  the  lengths  of  the 
parallel  sides.    Also  prove  that  it  is  parallel  to  the  parallel  sides. 

15.  In  what  ratio  does  the  point  (—  2,  .3)  divide  the  line  joining  the 
points  (-  3,  5)  and  (4,  -  9)  ?  Ans.  J. 

16.  In  what  ratio  does  the  point  (16,  3)  divide  the  line  joining  the 
points  (-  5,  0)  and  (2,  1)  ?  Ans.  -  I. 

17.  In  any  triangle  show  that  a  line  joining  the  middle  points  of  any    ^^ 
two  sides  is  parallel  to  the  third  side  and  equal  to  one  half  of  it. 


24 


NEW  ANALYTIC  GEOMETRY 


18.  If  (2,  1),  (3,  3),  (6,  2)  are  the  middle  points  of  the  sides  of  a  triangle, 
what  are  the  cooi'dinates  of  the  vertices  of  the  triangle  ? 

Ans.  (-1,2),  (5,0),  (7,  4). 

19.  Three  vertices  of  a  parallelogram  are  (1,  2),  (—  5,  —  3),  (7,  —  6). 
What  are  the  coordinates  of  the  fourth  vertex  ? 

Ans.  (1,  -  11),  (-  11,  5),  or  (13,  -  1). 

20.  The  middle  point  of  a  line  is  (6,  4),  and  one  end  of  the  line  is  (5,  7). 
What  are  the  coordinates  of  the  other  end  ?  Ans.  (7,  1). 

21.  The  vertices  of  a  triangle  are  (2,  3),  (4,  -  5),  (—  3,  —  6).  Find  the 
coordinates  of  the  point  v^here  the  medians  intersect  (center  of  gravity). 

14.  Areas.  In  this  section  the  problem  of  determining  the 
area  of  any  polygon,  the  coordinates  of  whose  vertices  are 
given,  will  be  solved.    We  begin  with  the 

Theorem.  Tke  area  of  a  triangle  whose  vertices  are  the  origin, 
P^(x^,  y^,  and  P^{x,-^,  y^  is  given  by  the  formula 

(V)  Area  of  A  OPJ*^  =  |  {x^y^  -  x^y^) . 

Proof.    In  the  figure' let 

a  =  Z.  XOP^, 
P  (Greek  ''  beta")  =  Z  XOP.^, 
e  (Greek  "  theta  ")  =  Z  P^OP^. 

(1)  .-.e^p-a. 

By  45,  p.  4, 

(2)  Area  A  OP^P^  =  ^0P^-  OP^ sin  0 

=  iOP^.OP^sm((3-a)  (by  (1)) 

(3)  =  i  OP^  ■  OP^ (sin  ^  cos  a  —  cos  /3  sin  a). 

But  in  the  figure 


sma 


M,    X 


e 

(By  34,  p.  3) 

M,P-2__     ?/2 

OP,       OP,' 

o       OM,        X, 

M,Pi_    y, 
OP,        OP, ' 

OMi        Xi 

''''''=  oprop- 

Substituting  in  (3)  and  reducing,  we  obtain 
Area  A  OPJ^.^  =  ^{x^y^  -  x^y^). 


Q.E.D. 


CARTESIAN  COORDINATES 


25 


(-2;4) 

Y\ 

/ 

\ 

/ 

\ 

/ 

\ 

(1,1) 

y 

f 

l\ 

III 

L 

— ■ 

■ — 

'' 

0 

X 

(-6, 

-1) 

\ 

EXAMPLE 

Find  the  area  of  the  triangle  whose  vertices  are  the  origin,  (—2,  4), 
and  (—5,  —  1). 

Solution.    Denote  (-  2,  4)  by  P^,  (-  6,  -  1) 
by  Fi.   Tlien 

x,  =-2,     2/i  =  4,     ;c.,  =  — 6,     i/.,^— 1. 
Substituting  in  (V), 

Area  =  |  [- 2  •  -  1  -(- 5)  •  4]  =  11. 

Then  area  =11  tmii  squares. 

If,  however,  the  fonnula  (V)  is  applied  by  denoting  (—  2,  4)  by  Pg, 
and  (—5,  —  1)  by  P,.,  the  result  will  be  —  11. 

The  two  figures  for  this  example  are  drawn  below. 

The  cases  ot positive  and  negative  area  are  distinguished  by  the 

Theorem.  Passing  around  the 
perimeter  in  the  order  of  the 
vertices  0,  P^,  P^, 

if  the  area  is  on  the  left,  as  v/i. 
Fig.l,  then  (V)  gives  a  posi- 
tive result ; 

if  the  area  is  on  the  right,  as 

in  Fig.  2,  then  (V)  gives  a  negative  result. 

Proof.    In  the  formula 

(4)  Area  A  OPJ\^  =  |  OP^  ■  0I\  sin  9 

the  angle  6  is  measured  from  OP^  to  OP,,  irithi/i.  the  triangle. 
Hence  6  is  positive  when  the  area     p^  p 

lies  to  the  left  in  passing  around 
the  perimeter  0,  J\,  P^,  as  in  Fig.  1, 
since  0  is  then  measured  counter- 
clockwise (p.  2).  But  in  Fig.  2,  6  is 
measured  clockwise.  Hence  6  is  negative  and  sin  0  in  (4)  is 
also  negative.  q.e.D. 

We  apply  (V)  to  any  triangle  by  regarding  its  area  as  made 
up  of  trinngles  with  the  origin  as  a  common  vertex. 


Fig.  1 


Fig.  2 


Fig.  1 


26  NEW  ANALYTIC  GEOMETRY 

Theorem.    The  area  of  a  triangle  whose  vertices  are  P^(x^,  y^, 

(VI)  Area  A  P^P^P^  =  |  (x^y^  —  x^y^  +  x^y^  —  x^y^  +  x^y^  —  x^y^). 

This  formula  gives  a  jwsltlve  or  negative  result  according  as 
the  area  lies  to  the  left  or  right  In  2)asslng  around  the  perimeter 
in  the  order  P,  P„  P,.  j>  /> 

Proof    Two  cases  must  be  distin-  [C^^-'/         f^     /\ 

guished  according  as    the  origin   is  '^V/  t       //C  I 

within  or  Avithout  the  triangle.  ;/  '    /  ^^'-'^^ 

Fig.   1,  origin  within  the  triangle.  p  p 
By  inspection,                                                    Fig.  1  Fig.  2 

(5)  Area  A  P^P.,P^  =  A  OP.P^  +' A  OP^P^  +  A  OP^P^, 

since  these  areas  all  have  the  same  sign. 

Fig.  2,  origin  without  the  triangle.    By  inspection, 

(6)  Area  A  PjP^^g  =  A  OP^P,^  +  A  OP,,P^  +  A  OP^P^, 

since  OP^P.^,  OPJ\  have  the  sr?w-e  sign,  but  OP.^P^  the  opposite 
sign,  the  algebraic  sum  giving  the  desired  area. 

By  (V),  AOP^P,  =  i(x,y.3  -  ,T^.y^), 

and  AOP^P^  =  ^(x^g^-x^g.;). 

Substituting  in  (5)  and  (6),  we  have  (VI). 

Also  in  (5)  the  area  is  positive,  in  (6)  negative.  q.  e.  d. 

An  easy  way  to  apply  (VI)  is  given  by  the  following 

"Rule  for  findmg  the  area  of  a  triangle. 

First  step.  Write  down  the  vertices  In  two  colmnns, 
abscissas  in  one,  ordlnates  In  the  other,  repeating  the 
coordinates  of  tlie  first  vei'tex. 

Second  step.  3Iultlpli/  each  abscissa  by  the  ordinate  of  the 
next  row,  and  add  results.    This  gives  x^y^  +  3?oy3'+  ^^Ui- 

Third  step.  Multiply  each  ordinate  by  the  abscissa,  of  the  next 
row,  and  add  results.    This  gives  y^x,^  +  y,^x^  4-  y.^Xy 


^x 

Vi 

^2 

y^ 

^3 

?/8 

^1 

y^ 

CARTESIAN  COORDINATES 


27 


Fourth  step.  Subtract  the  result  of  the  third  step  from  that 
of  the  second  step,  and  divide  by  2.  This  gives  the  required 
area,  namely  formula  (VI). 

Formula  (VI)  may  be  readily  memorized  by  remarking  that 
the  right-hand  member  is  a  determinant  of  simple  form,  namely 

Area  A  P^P.f"^ 


In  fact,  when  this  determinant  is  expanded  by  the  usual 
rule,  the  result,  when  divided  by  2,  is  precisely  (VI). 

It  is  easy  to  show  that  the  above  rule  applies  to  any  polygon 
if  the  following  caution  be  observed  in  the  first  step : 

Write  down  the  coordinates  of  the  vertices  in  an  order  agreeing 
with  that  established  bg  passing  continuously  aro^ind  the  pjerim- 
eter,  and  repeat  the  coordinates  of  the  first  vertex. 


Area 


25  +  25 


=  26  unit  squares.     Ans. 


The  result  has  the  positive  sign,  since  the 
area  is  on  the  left. 


1 

-1 

-3 

2 

1 


EXAMPLE 

Find  the  area  of  the  quadrilateral  whose  vertices  are  (1,  6),  (—  3, 
(2,  -2),  (-1,3). 

Solution.  Plotting,  we  have  the  figure  from  which  we  choose 
the  order  of  the  vertices  as  indicated  by  the  arrows.  Following 
the  rule : 

First  step.    Write  down  the  vertices  in  order. 

Second  step.  Multiply  each  abscissa  by  the 
ordinate  of  the  next  row,  and  add.    This  gives 

1  X  3  +  (-  1  X  -  4)  +  (-  3  X  -  2)  +  2  X  6  =  2.5. 

Third  step.  Multiply  each  ordinate  by  tlie 
abscissa  of  the  next  row,  and  add.  This  gives 
G  X  -  1  +  3  X  -  3  +  (- 4  X  2)  +  (- 2  X  1)  =  -  25. 

Fourth  step.  Subtract  the  result  of  the  third 
step  from  the  result  of  the  second  step,  and 
divide  by  2 


6 

3 

-  4 

_  2 

6 


Y 

(1, 

/ 

1 

r 

i' 

/ 

I 

± 

^ 

/ 

I 

It 

/ 

\\ 

/ 

y 

' 

0 

X 

^ 

-^ 

(2,- 

2) 

^ 

« 

.-4) 

28  NEW  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3),  (1,  5), 
({-  1,  -2).  Anf<.    V-. 

2.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3),  (4,  —  5), 
,(_  3,  _  0).  Anfi.    29. 

3.  Find  the  area  of  the  triangle  whose  vertices  are  (8,  3),  (—  2,  3), 
(4,  -  5).  Ans.    40. 

4.  Find  the  area  of  the  triangle  whose  vertices  are  (rt,  0),  (—  a,  0), 
i(0,  b).  Ans.    ab. 

5.  Find  the  area  of  the  triangle  whose  vertices  are  (0,  0),  (x,,  y^), 

2 

6.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  1),  (0,  b),  (c,  1). 

Ans.  (^LzJ^lfc^i). 
2 

7.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  6),  (6,  «), 
(c,  -c).  Am.    i(a2-6'-2). 

8.  Find  the  area  of  the  triangle  whose  vertices  are  (3,  0),  (0,  3\'^), 
(6,  3V3).  Ans.    9  Vs. 

9.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(2,  3),  (5,  4),  (—  4, 1)  is  zero,  and  hence  that  these  points  all  lie  on  the 
same  straight  line. 

10.  Prove  that  the  area  of  tlie  triangle  whose  vertices  are  the  points 
(a,  b  -\-  c),  (6,  c  +  a),  (c,  a  +  b)  is  zero,  and  hence  that  these  points  all  lie 
on  the  same  straight  line. 

11.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(a,  c  +  a),  (—  c,  0),  {—  a.,  c  —  a)  is  zero,  and  hence  that  these  points  all 
lie  on  the  same  straight  line. 

12.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (—  2,  3), 
(_  3,  _  4),  (  5,  -  1),  (2,  2).  Ans.    31. 

13.  Find  the  area  of  the  pentagon  whose  vertices  are  (1,  2),  (3,  —  1), 
(6,  -  2),  (2,  5),  (4,  4).  Ans.    18. 

14.  Find  the  area  of  the  parallelogram  whose  vertices  are  (10,  5), 
(-2,5),  (-5,  -3),  (7,  -3).  Ans.    96. 

15.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (0,  0),  (5,  0), 
(9,11),  (0,  3).  ^ns.    41. 


CARTESIAN  COORDINATES  29 

16.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (7,  0),  (11,  9), 
(0,  5),  (0,  0).  Ans.    59. 

17.  Show  that  the  area  of  the  triangle  whose  vertices  are  (4,  6),  (2,  —  4), 
(—  4,  2)  is  four  times  the  area  of  the  triangle  formed  by  joining  the 
middle  points  of  the  sides. 

18.  Show  that  the  lines  drawn  from  the  vertices  (3,  —  8),  (—  4,  6), 
(7,  0)  to  the  point  of  intersection  of  the  medians  of  the  triangle  divide 
it  into  three  triangles  of  equal  area. 

19.  Given  the  quadrilateral  whose  vertices  are  (0,  0),  (6,  8),  (10,  —  2), 
(4,  —  4);  show  that  the  area  of  the  quadrilateral  formed  by  joining  the 
middle  points  of  its  adjacent  sides  is  equal  to  one  half  the  area  of  the 
given  quadrilateral. 


CHAPTER  III 

CURVE  AND  EQUATION 

15.  Locus  of  a  point  satisfying  a  given  condition.  The  curve* 
(or  group  of  curves)  passing  through  all  points  which  satisfy 
a  given  condition,  and  through  no  other  points,  is  called  the 
locus  of  the  point  satisfying  that  condition. 

For  example,  in  plane  geometry,  the  following  results  are 
proved : 

The  perpendicular  bisector  of  the  line  joining  two  fixed 
points  is  the  locus  of  all  points  equidistant  from  these  points. 

The  bisectors  of  the  adjacent  angles  formed  by  two  lines  are 
the  locus  of  all  points  equidistant  from  these  lines. 

To  solve  any  locus  problem  involves  two  things  : 

1.  To  draw  the  locus  by  constructing  a  sufficient  number  of 
points  satisfying  the  given  condition  and  therefore  lying  on 
the  locus. 

2.  To  discuss  the  nature  of  the  locus;  that  is,  to  determine 
properties  of  the  curve. 

Analytic  geometry  is  peculiarly  adapted  to  the  solution  of 
both  parts  of  a  locus  problem. 

16.  Equation  of  the  locus  of  a  point  satisfjring  a  given  condition. 
Let  us  take  up  the  locus  problem,  making  use  of  coordinates. 
We  imagine  the  point  P{x,  y)  vioving  in  such  a  manner  that 
the  given  condition  is  fulfilled.  Then  the  given  condition  will 
lead  to  an  equation  involving  the  variables  x  and  y.  The 
following  example  illustrates  this. 

*The  word  "curve"  will  hereafter  signify  any  continuous  line,  straight 
or  curved. 

30 


CURVE  AND  EQUATION 


31 


EXAMPLE 

The  point  P  (x,  y)  moves  so  that  it  is  always  equidistant  from 
A  (—  2,  0)  and  B{—  3,  8).    Find  the  equation  of  the  locus. 

Solution.  Let  P{x,  y)  be  any  point  on  the  locus.  Then  by 
the  given  condition 

(1)  PA  =  PB. 

But,  by  formula  (I),  p.  13, 


PA  =  ^(x-\-2y  +  (y-0y, 
and    PB  =  -s/(x  +  Sf  +  (y  -  Sf. 
Substituting  in  (1), 

(2)  V(:r  +  2)^  +  (y-0)2 
=  V(.;  +  3)2  +  (y-8)l 

Squaring  and  reducing, 

(3)  2a; -16?/ +  69 


0. 


In  the  equation  (3),  x  and  ?/  are  variables  representing  the 
coordinates  of  any  point  on  the  locus ;  that  is,  of  any  point 
on  the  perpendicular  bisector  of  the  line  AB.  This  equation 
is  called  the  equation  of  the  locus  ;  that  is,  it  is  the  equation  of 
the  perpendicular  bisector  CP.  It  has  two  important  and 
characteristic  properties : 

1.  The  coordinates  of  any  point  on  the  locus  may  be  sub- 
stituted for  X  and  y  in  the  equation  (3),  and  the  result  will  be 
true. 

For  let  P^{x^,  y^)  be  any  point  on  the  locus.  Then  P^A  =  P^B, 
by  definition.    Hence,  by  formula  (I),  p.  13, 


(4)  V(a;^  +  2y  +  yl  =  ^(x^  +  Sf  +  (y,  -  Sf, 
or,  squaring  and  reducing, 

(5)  2  a;^- 16//, +  69  =  0. 


32  NEW  ANALYTIC   GEOMETRY 

But  this  equation  is  obtained  by  substituting  x^  and  ?/j  for  x 
and  1/  respectively  in  (3).    Therefore  x^  and  i/^  satisfy  (3). 

2.  Conversely,  every  point  whose  coordinates  satisfy  (3)  will 
lie  upon  the  locus. 

For  if  P^i^Cp  y^  is  a  point  whose  coordinates  satisfy  (3),  then 
(5)  is  true,  and  hence  also  (4)  holds.  q.e.  d. 

In  particular,  the  coordinates  of  the  middle  point  C  oi  A 
and  B,  namely,  a:=— 2J^,  y/  =  4  (IV,  p.  21),  satisfy  (3),  since 
2(-2i)-16  X  4  +  69  =  0. 

This  discussion  leads  to  "the  definition : 

The  equation  of  the  locus  of  a  point  satisfying  a  given  condi- 
tion is  an  equation  in  the  variables  x  and  y  representing  coor- 
dinates such  that  (1)  the  coordinates  of  every  point  on  the 
locus  will  satisfy  the  equation  ;  and  (2)  conversely,  every  point 
whose  coordinates  satisfy  the  equation  will  lie  upon  the  locus. 

This  definition  shows  that  the  equation  of  the  locus  must  be 
tested  in  two  ivays  after  derivation,  as  illustrated  in  the  exam- 
ple of  this  section.  The  student  should  supply  this  test  in  the 
examples  and  problems  of  Art.  17. 

From  the  above  definition  follows  at  once  the 

Corollary.  A  point  lies  upoii  a  curve  ivhen  and  only  lahen  its 
coordinates  satisfy  the  equation  of  the  curve. 

17.  First  fundamental  problem.  To  find  the  equation  of  a 
curve  which  is  dejined  as  the  locus  of  a  point  satisfying  a  given 
condition. 

The  following  rule  will  suffice  for  the  solution  of  this  prob- 
lem in  many  cases  : 

Rule.  First  step.  Assume  that  P(x,  y)  is  any  jjoint  satisfying 
the  given  condition,  and  is  therefore  on  the  curve. 

Second  step.    Write  down  the  given  condition. 

Third  step.  Express  the  given  condition  in  coordinates  and 
simplify  the  result.  The  final  equation,  containing  x,  y,  and  the 
given  constants  of  the  problem,  will  he  the  required  equation. 


CURVE  AND  EQUATION 


33 


EXAMPLES 


1.  Find  the  equation  of  the  straight  line  passing  through  P^  (4,  —  1) 

and  having  an  inclination  of 

4 

Solution.    First  step.    Assume  P  (x,  y)  any  point  on  the  line. 

Second  step.    The  given  condition,  since  the 

.     ,.       .  .    Sir 

inclination  a:  is  —  ,  may  be  written 

(1)  slope  of  P^P  =:  tan  or  =  —  1. 
Third  step.    From  (11),  p.  17, 

(2)  slope  of  P^P  =  tan  a  =  hJlJ^  _  L+i  . 
[By  substituting  (x,  y)  for  (x^,  y-^),  and  (4,  —  1)  for  (X2,  y^)-] 

y  +  i 


Therefore,  from  (1), 


(3) 


—  1,  or 

X  —  4 

X  +  y  —  3  =  0.    Ans. 


2.  Find  the  equation  of  a  straight  line  parallel  to  the  axis  of  y  and  at 
a  distance  of  6  units  to  the  right. 

Solution.  First  step.  Assume  that  P  (x,  ?/) 
is  any  point  on  the  line,  and  draw  NP  per- 
pendicular to  OY. 

Second  step.  The  given  condition  may  be 
written 

(4)  NP  =  6. 

Third  step.  Since  NP  =  OM  =  x,  (4)  be- 
comes 

(5)  X  =  6.    Ans. 

3.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from 
(—1,2)  is  always  equal  to  4. 

Solution.     First  step.    Assume  that  P  (x,  y)  is  any  point  on  the  locus. 

Second  step.    Denoting  (—  1,  2)  by  C,  the  given  condition  is 

(6)  PC  =  4. 


}'i 

1. 

:v' 

._. 

— , 

__. 

._. 

_.. 

p 

(X 

iv) 

0 

M 

X 

34 


NEW  ANALYTIC  GEOMETRY 


Third  step.    By  formula  (I),  p.  13, 


PC  =  V(x+l)2  +  {2/-2)2 
Substituting  in  (6), 


Squaring  and  reducing, 
(7)       x2^  2/2  + 2a; -42/ -11  =  0. 

This  is  the  required  equation,  namely, 
the  equation  of  the  circle  whose  center  is 
(—  1,  2)  and  radius  equal  to  4. 


}', 

k. 

.^ 

^ 

"^ 

s. 

/ 

/ 

> 

< 

{Ai> 

/ 

'' 

\l 

C 

,' 

K 

-1,2) 

V 

/ 

N 

\^ 

0 

y 

/ 

X 

s 

•^^ 

^ 

/ 

1 

PROBLEMS 

1.  Find  the  equation  of  a  line  parallel  to  OY  and 

(a)  at  a  distance  of  4  units  to  the  right. 

(b)  at  a  distance  of  7  units  to  the  left. 

(c )  at  a  distance  of  2  units  to  the  right  of  (3,  2) . 

(d)  at  a  distance  of  5  units  to  the  left  of  (2,  —  2). 

2.  Find  the  equation  of  a  line  parallel  to  OX  and 

(a)  at  a  distance  of  3  units  above  OX. 

(b)  at  a  distance  of  6  units  below  OX. 

(c)  at  a  distance  of  7  units  above  (—  2,  —  3). 

(d)  at  a  distance  of  5  units  below  (4,  —  2). 

3.  What  is  the  equation  of  XX'  ?  of  YY'  ? 

4.  Find  the  equation  of  a  line  parallel  to  the  line  x  =  4  and  3  units 
to  the  right  of  it ;   8  units  to  the  left  of  it. 

5.  Find  the  equation  of  a  line  parallel  to  the  line  y  =  —  2  and  4  units 
below  it ;   5  units  above  it. 

6.  What  is  the  equation  of  the  locus  of  a  point  which  moves  always 
at  a  distance  of  2  units  from  the  axis  of  x  ?  from  the  axis  oi  y?  from 
the  line  x  =  —  5  ?   from  the  line  ?/  =  4  ? 

7.  What  is  the  equation  of  the  locus  of  a  point  which  moves  so  as  to 
be  equidistant  from  the  lines  x  =  5  and  x  =  9  ?  equidistant  from  y  =  3 
and  y  =—7? 

8.  What  are  the  equations  of  the  sides  of  the  rectangle  whose  vertices 

are  (5,  2),  (5,  5),  (-2.  2).  (-2.  F,\  ? 


CURVE  AND  EQUATION  35 

In  Problems  9  and  10,  P^  is  a  given  point  on  the  required  line,  rn  is 
the  slope  of  the  line,  and  a  its  inclination. 

9.  What  is  the  equation  of  a  line  if 

(a)  Pj  is  (0,  3)  and  m=-3?  Ans.   3x  +  y  -  S  =  0. 

(b)  Pj  is  (-  4,  -  2)  and  7n  =  i?         Ans.   x-Sy-2  =  0. 

V2 

(c)  P^  is  (- 2,  3)  and  7n  =  ^^  ?         j^^s.   V2x  -  2?/ +6  +  2  V2  =  0. 

y/3  " 

(d)  Pi  is  (0,  5)  and  ?n  =  -—  ?  ^ns.   V3  x  -  2  y  +  10  =  0. 

(e)  P^  is  (0,  0)  and  m  =  -  |  ?  Anfi.    2  x  +  3  «/  =  0. 

(f)  P,  is  (a,  b)  and  77i  =  0?  Ans.    y  =  b. 

(g)  P^  is  (—  a,  b)  and  m  =  oo  ?  Ans.    z  =—  a. 

10.  What  is  the  equation  of  a  line  if 

(a)  Pi  is  (2,  3)  and  a  =  45°?  Ans.    x  -  y  +  1  =  0. 

(b)  Pj  is  (-  1,  2)  and  a  =  45^ ?  Ans.  x  —  y  +  3  =  0. 

(c)  Pj  is  (—  a,  —  b)  and  or  =  45°  ?  ^ns.   x  —  y  =  b  —  a. 

(d)  Pj  is  (5,  2)  and  a  =  QO°?  Ans.   VSx  -  y  +  2-  bVs  =  0. 

(e)  P,  is  (0,  -  7)  and  a  -60°?  Ans.  VSx-  y—7  =  0. 

(f)  Pj  is  (-  4,  5)  and  a  =  0°?  Ans.   y  =  5. 

(g)  Pj  is  (2,  —  3)  a,nd  a  =  90°  ?  Ans.   x  =  2. 

(h)  Pj  is  (3,  -  3  V3)  and  a  =  120° ?    Ans.  ^3x  +  y^  0. 

(i)  Pj  is  (0,  3)  and  a=  150°  ?  Ans.   V3x  +  3y-9  =  0. 

(j)  P,  is  (a,  6)  and  a  =  135°?  Ans.    x  +  y  =  a  +  6. 

11.  Find  the  equation  of  the  strai;:,dit  line  which  passes  through  the 
points 

(a)  (2,  3)  and  {-  4,  -  5).  Ans.    4x-3!/4-l  =  0. 
Hint.   Find  the  slope  by  (II),  p.  17,  and  then  proceed  as  in  Problem  9. 

(b)  (2,  -  5)  and  (-  1,  9).  Ans.    14  x  +  3?/  -  13  =  0. 

(c)  (- 1,  6)  and  (G,  -  2),  Ans.    8x+72/-84  =  0. 

(d)  (0,  -  3)  and  (4,  0).  Ans.    3x-42/-12  =  0. 

(e)  (8,  -  4)  and  (-  1,  2).  Ans.    2x  +  3?/-4  =  0. 

12.  Find  the  equation  of  the  circle  with 

(a)  center  at  (3,  2)  and  radius  =  4.   Ans.    x^  ■\-  y'^  —  Qx  —  4y  —  3  —  0. 

(b)  center  at  (12,  —  6)  and  /•  =  13.       Ans.   x-  +  Z/"^  —  24x  +  lOy  =  0. 

(c)  center  at  (0,  0)  and  radius  —  r.  Ans.    x^  +  y-  =  r-. 

(d)  center  at  (0,  0)  and  r  =  5.  Ans.   x^  +  y^  =^  26. 

(e)  center  at  (3  a,  4  a)  and  r  =  5  a.    A ns.   x'^  +  y^  —  2  a  (3  x  +  4  y)  =  0. 

(f )  center  at  (6  4-  c,  6  —  c)  and  r  =  c. 

Ans.   x^  +  y^  -  2  {b  +  c)x  —  2  {b  —  c.)y  +  2 h^  +  c-  =  0. 


36  NP]W  ANALYTIC   GEOMETRY 

13.  Find  the  equation  of  a  circle  wliose  center  is  (5,  —  4)  and  whose 
circumference  passes  througli  the  point  (—  2,  3). 

14.  Find  the  equation  of  a  circle  having  the  line  joining  (.3,  —  5)  and 
(—  2,  2)  as  a  diameter. 

15.  Find  the  equation  of  a  circle  touching  each  axis  at  a  distance  6  units 
from  the  origin. 

16.  Find  the  equation  of  a  circle  whose  center  is  the  middle  point  of 
the  line  joining  (—  6,  8)  to  the  origin  and  whose  circumference  passes 
through  the  point  (2,  3). 

17.  A  point  moves  so  that  its  distances  from  the  two  fixed  points 
(2,  —  3)    and  (—  1,  4)  are  equal.    Find  the  equation  of  the  locus. 

Ans.   3x—7y  +  2  =  0. 

18.  Find  the  equation  of  the  perpendicular  bisector  of  the  line  joining 

(a)  (2,  1),  (- 3,  -  .3).  Ans.  lOx  +  8 y +13  =  0. 

(b)  (3,  1),  (2,  4).  Ans.  x  -  3?/  +  5  =  0. 

(c)  (-  1,  -  1),  (3,  7).  Ans.  x  +  2y-7  =  0. 

(d)  (0,  4),  (3,  0).  Ans.  6x-8y  +  7  =  0. 

(e)  (Xi,  ?/i),  (Xo,  ?y,). 

Ans.  2  (x^  —  X2)x  +  2  (y^  —  y^y  +  x^  -  x^  +  y^J  -  y^-  =  0. 

19.  Show  that  in  Problem  18  the  coordinates  of  the  middle  point  of  the 
line  joining  the  given  points  satisfy  the  equation  of  the  perpendicular 
bisector. 

20.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of  tlie 
triangle  (4,  8),  (10,  0),  (G,  2).    Show  that  they  meet  in  the  point  (11,  7). 

18.  Locus  of  an  equation.  The  preceding  sections  have  iUus- 
trated  the  fact  that  a  locus  problem  in  analytic  geometry  leads 
at  once  to  an  equation  in  the  variables  x  and  ?/.  This  equation 
having  been  found  or  being  given,  the  complete  solution  of  the 
locus  problem  requires  two  things,  as  already  noted  in  Art.  IS 
of  this  chapter,  namely  : 

1.  To  draw  the  locus  by  plotting  a  sufficient  number  of 
points  whose  coordinates  satisfy  the  given  equation,  and  through 
which  the  locus  therefore  passes. 

2.  To  discuss  the  nature  of  the  locus  ;  that  is,  to  determine 
properties  of  the  curve. 


CURVE  AND  EQUATION  37 

These  two  problems  are  respectively  called  : 

1.  Plotting  the  locus  of  an  equation  (second  fundamental 
problem). 

2.  Discussing  an  equation  (third  fundamental  problem). 
For  the  present,  then,  we  concentrate  our  attention  upon  some 

given  equation  in  the  variables  x  and  y  (one  or  both)  and  start 
out  with  the  definition  : 

The  locus  of  an  equation  in  two  variables  representing  coordi- 
nates is  the  curve  or  group  of  curves  passing  through  all  points 
whose  coordinates  satisfy  that  equation,*  and  through  such 
points  only. 

From  this  definition  the  truth  of  the  following  theorem  is  at 
once  apparent : 

Theorem  I.  If  the  form  of  the  given  equation  he  changed  in 
any  way  {for  example,  by  transposition,  by  multipUcation  by  a 
constant,  etc.),  the  locus  is  entirely  unaffected. 

We  now  take  up  in  order  the  solution  of  the  second  and  third 
fundamental  problems. 

19.  Second  fundamental  problem. 

Rule  to  plot  the  locus  of  a  given  equation. 

First  step.  Solve  the  given  equation  for  one  of  the  variables 
in  terms  of  the  other.f 

*  An  equation  in  the  variables  x  and  y  is  not  necessarily  satisfied  by  the 
coordinates  of  any  points.  For  coordinates  are  real  numbers,  and  the  form  of 
the  equation  may  he  such  that  it  is  satisfied  by  no  Teal  values  of  x  and  y.  For 
example,  the  equation  2.2  _(.  ^2  _|.  j  =  q 

is  of  this  sort,  since,  when  x  and  y  are  real  numbers,  x^  and  y^  are  necessarily 
positive  (or  zero),  and  consequently  X"+y^  +  1  is  always  a  positive  number 
greater  than  or  equal  to  1,  and  therefore  7iot  equal  to  zero.  Such  an  equation 
therefore  has  no  locus.  The  expression  "the  locus  of  the  equation  is  imagi- 
nary" is  also  used. 

An  equation  may  be  satisfied  by  the  coordinates  of  a,  finite  number  of  points 
only.  For  example,  a;-  +  ?/'^  =  0  is  .satisfied  by  a;  =  0,  ?/  =  0,  but  by  no  other  real 
values.  In  this  case  the  group  of  points,  one  or  more,  whose  coordinates  sat- 
isfy the  equation,  is  called  the  locus  of  the  equation. 

t  The  form  of  the  given  equation  will  often  be  such  that  solving  for  one  vari- 
able is  simpler  than  solving  for  the  other.   Always  choose  the  simpler  solution. 


38 


NEW  ANALYTIC  GEOMETRY 


Second  step.  By  this  formula  compute  the  values  of  the  vari- 
able for  which  the  equation  has  been  solved  by  assuming  real 
values  for  the  other  variable. 

Third  step.  Plot  the  points  corresponding  to  the  values  so 
determined. 

Fourth  step.  If  the  points  are  numerous  enough  to  suggest 
the  general  shape  of  the  locus,  draw  a  smooth  curve  through 
the  points. 

Since  there  is  no  limit  to  the  number  of  points  which  may 
be  computed  in  this  way,  it  is  evident  that  the  locus  may  be 
drawn  as  accurately  as  may  be  desired  by  simply  plotting  a 
sufficiently  large  number  of  points. 

Several  examples  will  now  be  worked  out.  The  arrangement 
of  the  work  should  be  carefully  noted. 


r^ 

k 

/, 

^ 

r 

y 

^ 

/ 

Y 

(0,2; 

Y 

/ 

y 

r-3,0j 

0 

\ 

A[ 

EXAMPLES 
1.  Draw  the  locus  of  the  equation 

2a;-3?/+6=:0. 
Solution.  First  step.  Solving  for  t/, 

Second  step.  Assume  values  for  x  and  com- 
pute ?/,  arranging  results  in  the  form  of  the 
accompanying  table  : 
Thus,  if 

X  =  1,  2/  =  5  •  1  +  2  =  2f, 

X  =  2,  2/  =  1 .  2  +  2  =  3i, 

etc. 

Third  step.    Plot  the  points  found. 

Fourth   step.    Draw   a   smooth    curve 

through  these  points. 

2   Plot  the  locus  of  the  equation 

2/  =  x2  -  2  X  -  3. 
Solution.  First  step.  The  equation  as  given  is  solved  for  y. 


X 

y 

X 

y 

0 

2 

0 

2 

1 

2t 

-  1 

1* 

2 

H 

—  2 

1 

3 

4 

-3 

0 

4 

n 

-4 

-1 

etc. 

etc. 

etc. 

etc. 

CURVE  AXD  EQUATION 


39 


Second  step.  Computing  y  by  assuming  values  of  x,  we  find  the  table  of 
values  below :  .,       y. 


X 

y 

X 

y 

0 

-3 

0 

-3 

1 

-4 

-  1 

0 

2 

-3 

-2 

5 

3 

0 

-3 

12 

4 

5 

-4 

21 

5 

12 

etc. 

etc. 

6 

21 

etc. 

etc. 

Third  step.   Plot  the  points. 
Fourth  step.    Draw  a  smooth  curve  through  these  points.    This  gives 
the  curve  of  the  fiirure. 


16  =  0. 


3.  Plot  the  locus  of  the  equation 

cc^  +  2/^  +  6  X 
Solution.  First  step.   Solving  for  ?/, 

y  -±  Vl6-6x-x2. 
Second  step.    Compute  y  by  assuming  values  of  x.   For  this  purpose  the 
table  of  Art.  3  will  be  found  convenient. 


X 

y 

X 

y 

0 

±4 

0 

±4 

1 

±3 

-  1 

±4.6 

2 

0 

-2 

±4.9 

3 

imag. 

-3 

±5 

4 

" 

-4 

±4.9 

5 

(; 

-  5 

±4.6 

6 

(; 

-  6 

±4 

7 

(( 

-  7 

±3 

-8 

0 

-9 

imag. 

For  example,  if  x  =  —  1,  r/  =  ±  Vl6  +0  —  1  =  V21  =  ±  4.6, 
if  X  =  3,  ?/  =  ±  Vl6  -  18  -  9  =  ±  V- 11, 
an  imaginary  number. 


40  NEW  ANALYTIC   GEOMETRY 

Third  step.    Plot  the  corresponding  points. 

Fourth  step.  Draw  a  smooth  curve  through  these  points. 

The  student  will  doubtless  remark  that  the  locus  of  Example  1,  p.  38, 
appears  to  be  a  straight  line,  and  also  that  the  locus  of  Example  3,  p.  39, 
appears  to  be  a  circle.  This  is,  in  fact,  the  case.  But  the  proof  must  be 
reserved  for  later  sections. 

PROBLEMS 

1.  Plot  the  locus  of  each  of  the  following  equations  : 

(a)  X  +  2  2/  =  0.  ( i )  X  =  2/2  +  2  2/  -  3.  (q)  x^  +  y^  =  25. 

(h)  x  +  2y  =  3.  (j)ix  =  y^.  {r)  x^  +  y^  +  9x  =  0. 

(c)  3x-2/ +  5  =  0.  {k)4x  =  ?/3-l.  (s)  x^  +  y^  +  4y  -  0. 

(d)z/  =  4x-2.  {l)y  =  x^-l.  (t)  x2  +  2/2-6x-16  =  0. 

(e)x2  +  4j/  =  0.  {m)y  =  x^-x.  (u)  x^  +  y^ —  6y -W  =  0. 

(f)2/  =  x2-3.  (n)  ?/  =  x3-x2-5.  (v)4?/  =  x*-8. 

(g)x2  +  4j/-5  =  0.  (o)x2  +  ?/2  =  4.  (w)4x  =  y*  +  8. 

(h)  ?/  =  x2  +  X  +  1.  (p)  x2  +  2/2  =  9.  (X)  4 2/2  =  x3  -  1. 

The  following  problem  illustrates  the 

Theorem.  If  an  equation  can  be  put  in  the  form  of  a  product  of  variable 
factors  equal  to  zero,  the  locus  is  found  by  setting  each  factor  equal  to  zero 
and  plotting  each  equation  separately. 

2.  Draw  the  locus  of  4  x2  —  9  2/^  =  0. 
Solution.   Factoring, 

(1)  (2x-32/)(2x  +  32/)  =  0. 

Then,  by  the  theorem,  the  locus  consists  of  the  straight  lines  (p.  59) 

(2)  2  X  -  3  y  =  0,  and 

(3)  2  X  +  3  2/  =  0. 

Proof.  1 .  The  coordinates  of  any  point  (x^ ,  2/  j )  which  satisfy  (1)  will  satisfy 
eit/i€r{2)  or  (3). 

For  if  (Xj,  2/1)  satisfies  (1), 

(4)  {2Xi-3  2/i){2x^  +  3  2/i)  =  0. 

This  product  can  vanish  only  when  one  of  the  factors  is  zero.  Hence 
either  2  Xj- 3  2/^  =  0, 

and  therefore  (Xj,  y^)  satisfies  (2) ; 
or  2Xi  +  32/i  =  0, 

and  therefore  (x^,  y^)  satisfies  (3). 


CURVE  AND  EQUATION  41 

2.  A  point  (Xj,  ?/j)  on  eitha'  of  the  lines  defined  by  (2)  and  (3)  will  also  lie 
on  the  locus  of  (l). 

For  if  (xi,  y^)  is  on  the  line  2  x  —  3  y  =  0,  then  (Corollary,  p.  32) 
(5)  2x1-32/1  =  0. 

Hence  the  product  (2  x^  —  3  ?/j)  (2  Xj  +  3  y.^)  also  vanishes,  since  by  (5) 
the  first  factor  is  zero,  and  therefore  (x^,  ?/,)  satisfies  (1). 

Therefore  every  point  on  the  locus  of  (1)  is  also  on  the  locus  of  (2)  and 
(3),  and  conversely.    This  proves  the  theorem  for  this  example.     Q.  E.  D. 

3.  Show  that  the  locus  of  each  of  the  following  equations  is  a  pair  of 
straight  lines,  and  plot  the  lines  : 

(a)  xV  =  0.  {\)  x^-  y-^  +  x  +  y  =  0. 

(b)  x2  =  9  2/2.  (m)  X-  -  3  xy  -4y^  =  0. 

(c)  x^  —  (/2  =  0.  {n  )  x'^  —  xy  +  5  X  —  5  y  —  0.         , 

(d)  ?/-  6?/ =  7.  (o)  x2-4?/2  +  .5x  +  10?/ =  0. 

(e)  xy  —  2x  =  0.  (p)  x^  +  2 xy  +  ?/  +  x  +  ?/  =  0. 

(f )  9 x2  -  ?/2  =  0.  {q)  x^  +  S xy  +  2y^_  +  X  +  y  =  0. 

(g)  x2-  ?jxy  =  0.  (r)  x2  — 2x7/+?/  +  Qx-  Gy  =  0. 
(h)  ?/  +  4x2/  =  0.  (s)  3x2  +  x?/  — 2?/2  +  6x-  4?/  =  0. 
(i)  x2-4x— 5  =  0.  (t)  3x2  — 2x?y  — 2/2  +  6x  — 52/ =  0. 
(j)  x?/— 2x2  — 3x  =  0.  (u)  x2  — 4x?/- 5?/  + 2x  — lOy  =  0. 
(k)  2/2- 5x?/+ G?/  =  0.  (v)  x2  +  4x?/  + 4  2/2+ 5x+10?/  + 6  =  0. 

4.  Show  that  the  locus  of  Ax-  +  2Jx  +  C  =  0  is  a  pair  of  parallel  lines, 
a  single  line,  or  that  there  is  no  locus  according  as  A  =  ^2  _  4  ^k;;  jg 
positive,  zero,  or  negative. 

5.  Show  that  the  locus  of  ^4x2  +  Bxy  +  Cy^  =  0  is  a  pair  of  intersect- 
ing lines,  a  single  line,  or  a  point  according  as  A  =  B'^  —  4  AC  is  positive, 
zero,  or  negative. 

6.  Show  that  the  following  equations  have  no  locus  (footnote,  p.  37) : 

(a)  X2  +  2/2  +  1  =0.  (e)  (x  +  l)2  +  ,/  +  4  =  0. 

(b)  2x2  +  3y-2  ^_  8.  (f)  x2  +  2/2  +  2x  +  2y  +  3  =  0. 

(c)  x2  4.  4  =  0.  (g)  4.x2  +  2/2  +  8x  +  5  =  0. 

(d)  x*  +  2/2  +  8  =  0.  (h)  2/*  +  2x2  +  4  =  0. 

(!)  9x2  +  4 7/2  +  I8x  +  82/  +  15  =  0. 

Hint.  Write  each  equation  in  the  form  of  a  sum  of  squares,  and  reason  as 
in  the  footnote  on  page  37. 


42  NEW  ANALYTIC   GEOMETRY 

20.  Third  fundamental   problem.    Discussion  of  an  equation. 

The  method  exphiined  of  solving  the  second  fundamental  prob- 
lem gives  no  knowledge  of  the  required  curve  except  that  it 
passes  through  all  the  points  whose  coordinates  are  determined 
as  satisfying  the  given  equation.  Joining  these  points  gives  a 
curve  more  or  less  like  the  exact  locus.  Serious  errors  may  be 
made  in  this  way,  however,  since  tlie  nature  of  the  curve  between 
any  two  successive  jyoints  j^lotted  is  not  determined.  This  objec- 
tion is  somewhat  obviated  by  determining  before  jdotting  cer- 
tain properties  of  the  locus  by  a  discussion  of  the  given  equation 
now  to  be  explained. 

The  nature  and  properties  of  a  locus  depend  upon  the  form 
of  its  equation,  and  hence  the  steps  of  any  discussion  must 
depend  upon  the  particular  problem.  In  every  case,  however, 
certain  questions  should  be  answered.  These  questions  will 
now  be  presented.  * 

1.  Is  the  curve  symmetrical  with  respect  to  either  axis  of  coor- 
dinates or  with  respect  to  the  origin?  , 

To  answer  this  question  we  may  proceed  as  in  the  following 
example  : 

EXAMPLE  '     ' 

Discuss  the  symmetry  of  the  locus  of 

(1)  .'c--f4  7/-=16. 

Solution.  The  equation  contains  no  odd  powers  of  x  ov  y  \ 
hence  it  may  be  written  in  any  one  of  the  forms 

(2)  (;xf  -(-  4  (-  7/)-  =  16,  replacing  {x,  y)  by  {x,  -  y) ; 

(3)  (-  x)""  +  4  {yf  =  16,  replacing  {x,  y)  by  (-  x,  y) ; 

(4)  (-  xf  -f  4  (-  yf  =  16,  replacing  (x,  y)  by  (-  x,  -  //). 

The  transformation  of  (1)  into  (2)  corresponds  in  the  figure 
to  replacing  each  point  P(x,  y)   on  the  curve   by  the   point 


CURVE  AND  EQUATION 


4a 


Q{x,  —  y).  But  the  points  P  and  Q  are  symmetrical  with 
respect  to  XX\  and  (1)  and  (2)  have  the  same  locus  (Theo- 
rem I,  p.  37).  Hence  the  locus 
of  (1)  is  unchanged  if  each 
point  is  changed  to  a  second 
point  symmetrical  to  the  first 
with  respect  to  A' A''.  There- 
fore the  loc2cs  is  si/ynmetrlcal 
with  respect  to  the  axis  of  x. 
Similarly,  frowi  .(3),  the  locus  is  symmetrical  with  respect  to 
the  axis  of  y,  and  from  (4)  the  locus  is  symmetricdl  with 
respect  to  the  origin,  for  the  points  P{x,  y)  and  S(^—x,  —  y) 
are   symmetrical  with  respect  to  the  origin,  since   OP  =  OS. 

In  plotting  the  equation  we  take  advantage  of  our  knowl- 
edge of  the  symmetry  of  the  curve  by  limiting  the  calcula- 
tion to  points  in  the  first  quadrant,  as  in  the 
table.'  We  plot  these  points,  mark  off  the  j^oints 
symmetrical  to  them  with  respect  to  the  axes 
and  the  origin,  and  then  draw  the  curve. 

The .  locus  is  called  an  ellipse. 

The  facts  brought  out  in  the  example  are 
stated  in 

Theorem  II.  Symmetry.  If  the  locus  of  an  equation  is  un- 
affected by  replacliKj  y  by  —  y  throughout  its  equation,  the  locus 
is  syvfiTTietrical  with  respect  to  the  axis  of  x. 

If  the  locus  is  unaffected  by  changing  x  to  —  x  tlirough- 
out  its  equation,  the  locus  is  symmetrical  with  respect  to  the 
axis  of  y. 

If  the  locus  is  unaffected  by  changing  both  x  and  y  to  —  x  and 
—  y  throughout  its  equation,  the  locus  is  syviinetrlcal  with 
respect  to  the  origin. 

These  theorems  may  be  made  to  assume  a  somewhat  differ- 
ent form  if  the  equation  is  algebraic  in  x  and  y.    The  locus  of 


X 

y 

4 
3.4 

2.7 

0 

0 

1 

^\ 

2 

44 


NEW  ANALYTIC   GEOMETRY 


an  algebraic  equation  in  the  variables  x  and  y  is  called  an 
algebraic  curve.     Then  from  Theorem  II  follows 

Theorem    III.     Symmetry   of  an   algebraic  curve.     If  no    odd 

jiowers  of  y  occur  in  an  equation,  the  locus  is  symmetrical  with 
respect  to  XX' ;  if  no  odd  po^vers  of  x  occur,  the  locus  is  sym- 
inetrical  with  respect  to  YY'.  If  every  term  is  of  even*  degree, 
or  every  term  oJ%dd  deff7-ee,  the  loctis  is  symmetrical  xoith  respect 
to  the  origin.     ', 

The  second  question  arises  from  the  following  considerations  : 

Coordinates  are  real  numbers.    Hence  all  values  of  a;  which 

give  imaginary  values  of  y  must  be  excluded  in  the  calculation. 

Similarly,  all  values  of  y  which  lead  to  imaginary  values  of  x 

must  be  excluded.    The  second  question  is,  then : 

2.    IVhat  values,  if  any,  of  either  coordinate  will  give  im,agi- 
nary  values  of  the  other  coordinate  ? 

The  following  examples  illustrate  the  method :  ^ 

EXAMPLES 

1.  What  values  of  x  and  y,  if  any,  must  be  excluded  in  determining 
points  on  the  locus  of 


(1) 


x2  +  4  2/2  =  16  ? 


X  =  ±  2  V4  —  y'^, 
y  =  ±  iVl6-x2. 


Solution.    Solving  for  x  in  terms 
of  y,  and  also,  for  y  in  terms  of  x, 

(2)  f  —  Ji  9  a/4  _  1/2 

(3) 

From  the  radical  in  (2)  -we  see 
that   all    values   of   y   numerically  greater   than   2    will   make   4  —  2/^ 
negative,  and  hence  make  x  imaginary.   Hence  all  values  of  y  greater 
than  2  or  less  than  —  2  must  be  excluded. 

Similarly,  from  the  radical  in  (3),  it  is  clear  that  values  of  x  greater 
than  4  or  less  than  —  4  must  be  excluded. 


*  A  constant  term  is  to  be  regarded  as  of  zero  (even)  degree,  as  16  iu  (1),  p.  42. 


CURVE  AND  EQUATION 


45 


Therefore  in  determining  points  on  the  locus,  we  need  assume  for  y 
values  only  between  0  and  2,  as  on  page  43,  or  values  of  x  between  0  and  4 
inclusive. 

A  further  conclusion  is  this  :  The  curve  lies  entirely  within  the  rec- 
tangle bounded  by  the  four  lines 

a;  =  4,     x=— 4,     y-2,     y-—2, 

and  is  therefore  a  closed  curve. 

2.  What  values,  if  any,  of  the  coordinates  are  to  be  •(^eluded  in  deter- 
mining the  locus  of  '/ 

(4)  2/^-4x4.  15  =  0? 

Solution.    Solving  for  x  in  terms  of  y,  and  also  for  y  in  terms  of  x, 

(5)  x=  i(1.5  +  2/2), 

(6)  y  =.  i  V4X-15. 

From  (5)  any  value  of  y  will  give  a  real  value  of  x.  Hence  no  values 
of  y  are  excluded. 


K 

X 

y 

31 

0 

4 

±1 

4f 

±2 

6 

±3 

7f 

±4 

10 

±5 

12f 

±6 

etc. 

etc. 

From  the  radical  in  (0)  all  values  of  x  for  which  4x  —  15  is  negative 
mu.st  be  excluded  ;  that  is,  all  values  of  x  less  than  3f . 

The  locus  therefore  lies  entirely  to  the  right  of  the  line  x  =  3|. 
Moreover,  since  no  values  of  y  are  excluded,  the  locus  extends  to  infinity, 
y  increasing  as  x  increases. 

The  locus  is,  by  Theorem  III,  symmetrical  with  respect  to  the  x-axis, 
and  is  called  a  parabola. 


46 


NEW  AXALYTIC   GEOMETRY 


- 

~ 

— 

— 

- 

- 

A 

X 

l> 

/ 

/ 

/^ 

/ 

v 

• 

/ 

V 

0 

V 

r 

/ 

f 

/ 

r 

r 

If 

t-.f 

; 

1 

_ 

— 

r 

_ 

_ 

_ 

3.  Determine  what  values  of  x  and  ?/,  if  any,  must  be  excluded  in 
determining  the  locus  of 

<7)  4  y  =  x3. 

Solution.  Solving  for  x  in  terms  of  y,  and  also  for  y  in  terms  of  x 

(8)  X  =  VTi], 

(9)  y  =  \x^. 

From  these  equations  it  appears  that  no  values 
of  either  coordinate  need  be  excluded. 

The  locus  is,  by  Theorem  III,  symmetrical 
with  respect  to  the  origin.  The  coordinates  in- 
crease together  ;  the  curve  extends  to  infinity 
and  is  called  a  cubical  parabola. 

The  method  illustrated  in  the  examples 
is  summed  up  in  the 

Rule  to  determine  all  values  of  x  and  y 
which  must  he  excluded. 

Solve  the  equation  for'  x  in  terms  of  y, 
and  from,  this  result  determine  all  values 

of  y  for  which  the  computed  value  of  x  ivould  he   imaginary. 
These  values  of  y  must  he  excluded. 

^olve  the  equation  f>r  y  in  terms  of  x,  and  from  this  result 
determine  all  values  of  x  for  which  the  computed  value  of  y 
would  he   imaginary.    These  values  of  x  mtist  he  excluded. 

In  determining  excluded  values  of  x,  and  y  we  obtain  also 
an  answer  to  the  question  : 

3.  Is  the  curve  a,  closed  curve,  or  does  it  extend  to  infinity? 

The  points  of  intersection  of  the  curve  with  the  coordinate 
axes  should  be  found. 

The  intercepts  of  a  curve  on  the  axis  of  x  are  the  abscissas  of 
the  points  of  intersection  of  the  curve  and  A'A''. 

The  intercepts  of  a  curve  on  the  axis  of  y  are  the  ordinates 
of  the  points  of  intersection  of  the  curve  and  YY^. 


CURVE  AND  EQUATION  47 

Rule  to  find  the  intercejjts. 

Substitute  y  =  0  and  solve  for  real  values  of  x.  This  gives  the 
intercepts  on  the  axis  of  x. 

Substitute  X  =■  0  and  solve  for  real  values  of  y.  This  gives  the 
intercepts  on  the  axis  of  y. 

The  proof- of  the  rule  follows  at  once  from  the  definitions. 
The  rule  just  given  explains  how  to  answer  the  question  : 

4.    What  are  the  intercepts  of  the  locus? 

In  particular,  the  locus  may  pass  through  the  origin,  in  which 
case  one  intercept  on  each  axis  will  be  zero.  In  this  case  the 
coordinates  (0,  0)  must  satisfy  the  equation.  When  the  equa- 
tion is  algebraic  we  have 

Theorem  IV.    The  locus  of  an  algebraic  eq^iation  passes  throvgh 
the  origin  when  there  is  no  constant  term  in  the  equation. 
The  proof  is  immediate. 

21.  Directions  for  discussing  an  equation.  Given  an  equation, 
the  following  questions  should  be  answered  in  order  before 
plotting  the  locus. 

X  Is  the  origin  on  the  locus  ?  , 

3-    What  are  the  intercepts  ? 
(D2>.  Is  the  locus  symmetrical  with  respect  to  the  axes  or  the 
origin  ? 

4.  What  values  of  x  and  y  must  I>e  excluded? 

5.  Is  the  curve  closed,  or  does  it  2^ass  off  indefinitely  far  ? 

Answering  these  questions  constitutes  what  is  called  a  general 
discussion  of  the  given  equation.  The  successive  results  should 
be  immediately  transferred  to  the  figure.  Thus  when  the  inter- 
cepts have  been  determined,  ynark  them  off  on  th,e  axes.  Indicate 
which  axes  are  axes  of  symmetry.  The  excluded  values  of  x 
and  y  will  determine  lines  parallel  to  the  axes  which  the  locus 
will  not  cross.    Draw  these  lines. 


48 


NEW  ANALYTIC  GEOMETRY 


EXAMPLE 

Give  a  general  discussion  of  the  equation 
(1)  x^-i2j^  +  16y  =  0. 

Draw  the  locus. 

Solution.    1.  Since  the  equation  contains  no  constant  term,  the  origin 
is  on  the  curve. 

2.  Putting  ?/  =  0,  we  find  a;  =  0,  the  intercept  on  the  axis  of  x.  Putting 
a;  =  0,  we  find  y  =  0  and  4,  the  intercepts  on  the  axis  of  y. 

Lay  off  the  intercepts  on  the  axes. 


>J 

s 

Y. 

X 

k 

V 

*«s 

*»• 

^ 

">> 

•v, 

^ 

^ 

^ 

, — 

^, 

--- 

.'/= 

4 

ro> 

i) 

,'/= 

2 

X 

' 

6 

"^ 

— ■ 

X 

^ 

^ 

-*' 

">». 

^ 

y^ 

^ 

y 

^ 

"^ 

* 

^\ 

r' 

^ 

3.  The  equation  contains  no  odd  powers  of  x  ;  hence  the  locus  is 
symmetrical  with  respect  to  YY' . 

4.  Solvino;  for  x, 


(2)  X  =  ±  2  V;/-  -  4  y. 

All  values  of  y  must  be  excluded  which  make  the  expression  beneath 
the  radical  sign  negative.  Now  the  roots  of  ?/2  —  4?/  =  0  are  y  =  0  and 
?/  =  4.  For  any  value  of  y  between  these  roots,  y-  —  4y  is  negative.  For 
example,  y  =  2  gives  4  —  8  =  —  4.  Hence  all  values  of  y  between  0  and 
4  must  be  excluded. 

Draw  the  lines  y  =  0  and  y  =  4.  The  locus  does  not  come  between 
these  lines. 

Solving  for  ?/, 

(3)  y  =  2±i  Vx^  +  16. 

Hence  no  value  of  x  is  excluded,  since  x^  +  16  is  positive  for  all 
values  of  x. 


CURVE  AND  EQUATION 


49 


5.  From  (3),  y  increases  as  x  increases,  and  tlie  curve  exte^ids  out 
indefinitely  far  from  botli  axes. 

Plotting  the  locus,  using  (2),  the  curve  is  found  to  be  as  in  the  figure. 
The  curve  is  a  hyperbola. 

Sign  of  a  quadratic.  In  the  preceding  example  it  became  necessary 
to  dfterniine  for  what  values  of  y  the  quadratic  expression  y'^  —  iy  in  (2) 
was  positive. 

The  fact  made  use  of  is  this  : 

If  the  sign  of  a  quadratic  expression  is  negative  (or  positive)  for  any 
one  value  of  the  unknown  taken  betv?een  the  roots,  it  is  also  negative 
(or  positive)  for  every  value  of  the  unknown  between  the  roots. 

This  is  easily  seen  graphically.    For  take  any  quadratic 

(4)  Ax-  +  Bx+  C. 
Plot  the  equation  " 

■  (.5)  y  =  Ax"  +  Bx+  C. 

The  locus  of '  (5)  will  be  a  parabola  turned  upward  if  A  is  positive, 
downward  if  A  is  negative  (see  Example  2,  p.  38).  The  intercepts  on  the 
,£-axis    will    be    the  . 

roots  of  (4).  The 
values    of    y    from 

(5)  will  clearly  all 
have  one  sign  for 
all  values  of  a>  be- 
tween the  intei'- 
cepts,  and  the  op- 
posite  sign    for    all 

other  values  of  x.  We  see,  then,  that  the  values  of  the  quadratic  (4)  will 
have  one  sign  for  all  values  of  x  taken  between  the  roots,  and  the  opposite 
sign  for  all  other  values. 

To  apply  this,  consider  the  locus  of 


(<i) 


Vg  —  5x  —  x^. 


What  values  of  x  must  be  excluded  ?  To  answer  this,  find  the  roots  of 
0— 5x  — a'"  =  0.  They  are  x=—Q  and  x  =  1.  Take  any  value  of  x 
between  these  roots,  for  example,  x  =  0.  When  x  =  0,  the  quadratic 
6  —  5x  —  x"  equals  6,  a  positive  number.  Hence  6  —  5x  —  x"-^  equals  a 
positive  number  for  all  values  of  x  between  the  roots  —  G  and  1.  Then 
the  quadratic  is  negative  for  all  other  values;  hence  we  must  exclude 
values  of  X  <  —  6  and  also  x  >  1. 


50  NEW  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Give  a  general  discussion  of  each  of  the  following  equations  and 
draw  the  locus.    Make  sure  that  the  discussion  and  the  figure  agree. 

i(a)x2-4y  =  0.  '  {n)  9y- -  x"  =  0. 

-(b)  ?/-4x  + 3  =  0.  (o)  9(/'- +  r' =  0. 

(c)  x'^  +  4?/-  l(i  =  0.  (p)  2x2/  +  3j'-  4  =  0. 

(d)  9x2  _,.  y-z  _  18  =  0.  (q)  x^  +  4x2/  +  S?/^  +  8  =  0. 
,  (e)  x2  — 42/2_iG  =  0.  (r)  x^  +  xy  +  y^  —  4  =  0. 

( f  )  x^  -  4  2/2  +  10  =  0.  (  s )  x2  +  2  xy  -  3  i/2  =  4. 

(g)  x2-?/2  +  4  =  0.  (t)  2xy-i>^  +  4x  =  0. 

( h )  x2  —  ?/  +  X  =  0.  ( u )  3  X-  —  y  +  X  =  0. 

1  ( i  )  xy  -  4  =  0.  ( V  )  4  (/•-  -  2  X  -  y  =  0. 

( j  )  Oy  +  x^  =  0.  (w)  x2  -  2/2  +  Gx  =  0. 

\  (k)  4x- 2/3  =  0.  (X)  x2  + 4  2/2  + 8z/  =  0. 

( 1  )  6x  -  2/^  =  0.  (y)  9x2  +  y-  +  I8x-6y  =  0. 

.(m)  5x  — 2/  +  2/"  =  0.  (z)  9x2  _  ^^2  ^_  jgx  +  62/  =  0. 

2.  Determine  the  general  nature  of  the  locus  In  each  of  the  following 
equations.  In  plotting,  assume  particular  values  for  the  arbitrary  con- 
stants, but  not  special  values  ;  that  is,  values  which  give  the  equation  an 
added  peculiarity.* 


=  2?-x. 
'  =  2ry. 


(a)  2/-  =  2  )/(.(•.  (f)  x2-  y 

(b)  x2  —  2  my  =  vi'^.  (g)  x2  +  y' 

lc\  ^  +  ^  -  1  (^^)  ^"^  +  ^' 

^     '    «2^    62  •  (i)    x"  +  y 

(d)  2x2/ =  a2.  '  (J)  "^^  = 

,  .   ^2      ^  _  (k)  a22/  =  x3. 

a^      h- 

The  loci  of  the  equations  (a)  to  (i)  in  Problem  2  are  all  of  the  class 
known  as  conies,  or  conic  sections,  —  curves  following  straight  lines  and 
circles  in  the  matter  of  their  simplicity.  These  curves  are  obtained  when 
cross  sections  are  taken  of  a  right  circular  cone.  Various  definitions  and 
properties  will  be  given  later.   A  definition  often  used  is  the  following : 

A  conic  section  is  the  locus  of  a  point  whose  distances  from  a  fixed 
point  and  a  fixed  line  are  in  a  constant  ratio. 

*  For  example,  in  (a)  and  (b)  )»  =  0  i.s  a  special  value.  In  fact,  in  all  these 
examples  zero  is  a  special  value  for  any  constant. 


CURVE  AND  EQUATION  51 

3.  Show  that  every  conic  is  represented  by  an  equation  of  tlie  second 
degree  in  z  and  y. 

Hint.  Take  71"  to  coincide  with  the  fixed  line,  and  di-aw  XX'  through  the 
fixed  point.   Denote  the  fixed  point  hy  (p,  0)  and  the  constant  ratio  by  e. 

Ans.  (1  —  e2)  ^2  +  y-2  _  2px  +  p^  =  0. 

4.  Discuss  and  plot  the  locus  of  the  equation  of  Problem  3  : 

(a)  when  e  =  1.   The  conic  is  now  called  a.  parabola  (see  p.  45). 

(b)  when  e  <  1.   The  conic  is  now  called  an  ellipse  (see  p.  43). 

(c)  when  e  >  1.   The  conic  is  now  called  a  hyperbola  (see  p.  48). 

5.  A  point  moves  so  that  the  sum  of  its  distances  from  the  two  fixed 
points  (3,  0)  and  (—  3,  0)  is  constant  and  equal  to  10.   What  is  the  locus  ? 

Ans.  Ellipse  16 x^  +  25 1/^  =  400. 

6.  A  point  moves  so  that  the  difference  of  its  distances  from  the  two 
fixed  points  (5,  0)  and  (—  5,  0)  is  constant  and  equal  to  8.  What  is  the 
locus  ?  Ans.  Hyperbola  9z-  —  16y^  =  144. 

7.  Find  the  equations  of  the  following  loci,  and  discuss  and  plot  them. 

(a)  The  distance  of  a  point  from  the  fixed  point  (0,  2)  is  equal  to  its 
distance  from  the  a;-axis  increased  by  2. 

(b)  The  distance  of  a  point  from  the  fixed  point  (0,  —  2)  is  equal  to 
its  distance  from  the  y-axis  increased  by  2. 

(c)  The  distance  of  a  point  from  the  origin  is  equal  to  its  distance 
from  tlie  ?/-axis  increased  by  2. 

(d)  The  distance  of  a  point  from  the  fixed  point  (2,  —  4)  is  equal  to  its 
distance  from  the a;-axis  increased  by  5.  Ans.   2y  =  x^  —  4x  —  5. 

(e)  The  distance  of  a  point  from  the  point  (3,  0)  is  equal  to  half  its 
distance  from  the  point  (6,  0). 

(f )  The  distance  of  a  point  from  the  point  (8,  —  4)  is  twice  its  distance 
from  the  point  (2,  —  1). 

(g)  One  third  of  the  distance  of  a  point  from  the  point  (0,  3)  is  equal  to 
its  distance  from  the  z-axis  increased  by  unity.   Ans.   x-  —  8  y-  —  24?/  =  0. 

(h)  The  distances  of  a  point  to  the  fixed  point  (—  1,  0)  and  to  the  line 
4x  —  5  =  0  are  in  the  ratio  f.  Ans.    Qx"^  +  25 y"  +  90a;  =  0. 

8.  Prove  the  statement :  If  an  equation  is  unaltered  when  x  and  y  are 
interchanged,  the  locus  is  symmetrical  with  respect  to  the  line  y  —  x. 

Make  use  of  this  result  in  drawing  the  loci  of  : 

{^)xy  =  4.     {h)  x^- +  xy  +  y-^  =  9.     {c)  x^  +  ij'^  :=  1 .     {d)xi  +  y^  =  1. 

22.  Asymptotes.  The  following  examples  elueidatt^  difficul- 
ties arising  frequently  in  drawing  the  locus  of  an  equation. 


52 


NEW  ANALYTIC   GEOMETRY 


EXAMPLES 


1.  Plot  the  locus  of  the  equation 
(1)  xy-2y-4  =  0. 

Solution.    Solving  for  y, 
4 


(2) 


y 


X 


AVe  observe  at  once,  if  x 


i2/ 


This  is  interpreted  thus :  The  curve  ap- 
proaches the  line  x  =  2  as  it  passes  off  to 
infinity.  In  fact,  if  we  solve  (1)  for  x  and 
write  the  result  in  the  form 

4 

x  =  2  +  -' 

y 

it  is  evident  that  x  approaches  2  a,s  y 
increases  indefinitely.  Hence  the  locus 
extends  both  upward  and  downward 
indefinitely  far,  approaching  in  each  case 
the  line  x  =  2.    The  vertical  line  x  =  2  is  called  a  vertical  asymptote. 


X 

y 

X 

y 

0 

_  2 

0 

_  2 

1 

-  4 

-1 

■t 

3 

H 

-8 

-2 

-1 

l| 

-  16 

-4 

—   I 

2 

00 

-5 

4 

2i 

16 

2^ 

8 

-10 

-i 

3 

4 

etc. 

etc. 

4 

2 

5 

1 

6 

1 

12 

0.4 

etc. 

etc. 

Va 

1 

h 

= 

\ 

\ 

V 

> 

s 

^ 

-^ 

^ — J 

— 

^ — 

;^ 

oo 

oo 

y 

tj 

, 

^ 

O 

f2,o; 

*Y 

> 

N 

S 

V 

\ 

— 

1 

1 

"   1 

In  plotting,  it  is  necessary  to  assume  values  of  x  differing 
from  2,  both  less  and  greater,  as  in  the  table. 


slightly 


CURVE  AND  EQUATION 


53 


From  (2)  it  appears  that  y  diminishes  and  approaches  zero  as  x  in- 
creases indefinitely.  The  curve  tlaerefore  extends  indefinitely  far  to  the 
right  and  left,  approaching  constantly  the  axis  of  x.  The  axis  of  x 
is  therefore  a  horizontal  asymptote.* 

This  curve  is  called  a  hyperbola. 

In  the  problem  just  discussed  it  was  necessaiy  to  learn  what  value  x 
approached  when  y  became  very  large,  and  also  what  value  y  approached 
when  X  became  very  large.  These  questions,  when  important,  are  usually 
readily  answered,  as  in  the  following  examples  : 


2.  Plot  the  locus  of 


V 


2x  +  3 


(Fig-  1) 


3x-4 

When  X  is  very  great,  we  may  neglect  the  .3  in  the  numerator  (2x4-3) 
and  the  —  4  in  the  denominator  (3x  —  4).    That  is,  when  x  is  very  large, 

V  =  —  =  -  •      Hence  y  =  -  is  a  horizontal  asymptote. 
^      3x      3  3  ^    ^ 

The  equation  shows  directly  that  3x  —  4  =  0  or  x  =  f  is  a  vertical 

asymptote.    Or  we  may  solve  the  equation  for  x,  which  gives 


4^  +  3 


Hence,  when  y  is  very  large,  x 


Fig.  2 


3.  The  locus  of 


y 


2x  +  3 


x2_3x  +  2 

is  shown  in  Fig.  2.    There  are  two  vertical  asymptotes,  a:  =  1  and  x  =  2, 
since  the  denominator  x^  —  3x  +  2  =  (x  —  1)  (x  —  2).    A  branch  of  the 

*  For  oblique  asymptotes,  that  is,  asymptotes  not  parallel  to  either  axis, 
see  Art.  66. 


54 


NEW  ANALYTIC   GEOMETRY 


curve- lies  between  these  lines.    Furthermore,  when  x  becomes  large,  we 

2  X      2 
may  write  the  equation  y  =  —  =  -  =  0.  Hence  the  x-axis 

X-       X 
is  a  horizontal  asymptote.    A  few  points  of  the  locus 
are  given  in  the  table.    Note  that  different  scales  are  used 
for  ordinates  and  abscissas. 


X 

y 

0 

3 

3 

2 

5 
—  5 

3 

2 

0 
-24 

1  3 
12 

1 

B 

The  determination  of  the  vertical  and  hori- 
zontal asymptotes  of  a  curve  should  be  added 
to  the  discussion  of  the  equation  as  outlined  in 

Art.  21. 

PROBLEMS 

riot  each  of  the  following,  and  determine  the  horizontal  and  vertical 
asymptotes : 

1.  (a)  xy  +  y-S  =  0. 

(b)  xy  +  X  +  3  =  0. 

(c)  2x(/  +  2x  +  3?/ 

(d)  X-  +  xy  +  8  =  0. 

2.  (a)  x^y  -5  =  0. 

(b)  x^y  —  y  -\-2x 


(e)  y  = 

(f)  y  = 

(g)  y  = 
(h)  y  = 
(i)  y  = 


5 


x2- 

-3x 

4X--2 

X-- 

-4 

X  - 

-3 

x+1 

X2- 

-4 

X-  - 

-1 

(X- 

-2)(x 

+  3) 

(x  +  2)(x-.3) 


(e) 

2x;/  +  4x-  6?/  +  3  =  0 

(f) 

?/2  +  2xi/-4  =  0. 

=  0. 

). 

(g)  X2/  +  X  +  2  y  -  3  =  0. 
(h)  X2/  +  2/-x2  +  2x  =  0. 

(c) 

X2/2-  4x+  6  =  0. 

=  0. 

(d) 

x^y  -  2/  +  8  =  0. 

(J)  y 

x2-4 

X^  -f  X 

?/2 

(0)  4x  =  ^^. 
2/   —  9 

(k)x 

2/2 

(P)i2x=  ^y  . 

^  '               3-2/2 

(l)x 

y-2 
2/-3 

fn)  v-Z'-'-^V 

^'^^  ^  -  I     x     j 

(m)  y 

_x2-l 
4-x2 

x"- 

X—  1 

(n)  y 

x2-3x 
x2  +  3x 

+  2 
+  2 

X2 

^-^y         X2-3X  +  2 

PROBLEMS  FOR  INDIVEDUAL  STUDY 

Discuss  fully  and  draw  carefully  the  following  loci : 

1.  2/2  -  4  X2/  +  x3  =  0.  5.  {y  -  x)2  -  x-  {a-  -  x^)  =  0. 

2.  2/2  -  2  X2/  -  2  x2  +  x3  =  0.  6.  (y  -  x2)2  _  (a2  -  x^)  =  0. 

3.  2/2  _  3-2  4.  a.4  _  0.  7.  (2/  -  x2)2  -  x2  {(l-  -  X"")  =  0. 

4.  (y  _  a;)2  _  (a2  _  3.2)  _  q.  8.  ?/  («  _  a;)  _  ^3  =  q  (the  cissoid). 

9.  2/2  (a  —  x)  —  x2  (a  +  x)  =  0  (the  strophoid). 
10.  x*  +  2  ax^y  -  ay^  =  0. 


CURVE  AND  EQUATION  55 

11.  X*  —  axy'^  +(/4  =  0. 

12.  a*2/2  —  a^x*  +  x^  =  o. 

13.  02/2  _  bz*  -  x6  =  0. 

14.  a^i/2  _  2  a6x^?/  —  x^  =  0. 

15.  2/2  -  (a2  -  x2)  (62  -  x2)2  =  0. 

16.  x8y2  _  (fij.2  ^  aij*  =  0. 

17.  X  (2/  -  x)2  -  h^y  =  0. 

18.  (x2  +  2/2)2  _  ^2  (j;2  _  y2)  ^  0  (the  lemniscate). 

19.  (x2  -  a2)2  =  02/2  (3  a  +  2  y). 

20.  (x2  +  ,/-2  _  1)  y  _  ax  =  0. 

21.  2/2  -  x2  —  X  (x  —  4)2  =  0. 

22.  (x2  +  2/2-2  a2/)2  =  a2  (x2  +  y^)  (the  limacon). 

23.  (X*  +  x22/2  +y*)  =  x  (ax2  —  4  ay^). 

24.  (X2  +  2/2  ^  4  oty  _  (j2)  (j.2  _  q2)  ^  4  g[2y2  =  Q. 

25.  (2/2  -  x2)  (x  -  1)  (x  -  I)  =  2  (2/2  +  x2  -  2  x)2. 

26.  (x2  +  ,/2  -I-  4  a2/  -a^)  (x2  _  a2)  +  4  ahj-  =  0  (the  cocked  hat). 

23.  Points  of  intersection.  If  two  curves  whose  equations 
are  given  intersect,  the  coordinates  of  each  point  of  intersection 
must  satisfy  both  equations  when  substituted  in  them  for  the 
variables.  In  algebra  it  is  shown  that  all  values  satisfying 
two  equations  in  two  unknowns  may  be  found  by  regarding 
these  equations  as  simultaneous  in  the  unknowns  and  solving. 
Hence  the 

Rule  to  find  the  points  of  intersection  of  tivo  curves  whose 
eqiiatlons  are  given. 

Consider  the  equations,  as  simultaneous  in  tlie  coordinates 
and  solve  ms  in  algebra. 

Arrange  the  real  solutions  in  corresponding  jJf^i^'s.  These  will 
be  the  coordinates  of  all  tlie  points  of  intersection. 

Notice  that  only  real  solutions  correspond  to  common  points 
of  the  two  curves,  since  coordinates  are  always  real  numbers. 


56 


NEW  ANALYTIC   GEOMETRY 


EXAMPLES 

1.  Find  the  points  of  intersection  of 
<1)   X- 72/ +  25  =  0, 

(2)  x2  +  ?/2  =  25, 

Solution.  Solving  (1)  for  x, 

(3)  x=72/-25. 

Substituting  in  (2), 
{7y-  25)-^  +  y^  =  25. 

Reducing, 
2/2-72/4-12  =  0. 

.-.  y  —  S  and  4. 

Substituting  in  (3)  [not  in  (2)], 

X  =  —  4  and  +  3. 

Arranging,  the  points  of  intersection  are  (—  4,  3)  and  (3,  4).   Ans. 
In  the  figure  the  straight  line  (1)  is  the  locus  of  equation  (1),  and  the 
circle  the  locus  of  (2). 

2.  Find  the  points  of  intersection  of  the  loci  of 

(4)  2x2  +  32/2  =  35, 

(5)  3x2-42/  =  0. 
Solution.    Solving  (5)  for  x^, 

(6)  x2  =  |2/. 
Substituting  in  (4)  and  reducing, 

92/2  +  8  2/ -105  =  0. 

.-.  2/  =  3  and  —  ^. 
Substituting  in  (6)  and  solving, 
X  =  ±  2  and  ±  i  V- 210. 

Arranging  the  real  values,  we  find  the  points  of  intersection  are 
(4-  2,  3),  (-  2,  3).   Ans. 

In  the  figure  the  ellipse  (4)  is  the  locus  of  (4),  and  the  parabola  (5)  the 
locus  of  (-5). 


CURVE  AND  EQUATION  67 

PROBLEMS 
Find  the  points  of  intersection  of  the  following  loci  : 
7x-Uy  +  l=0^  x^  +  y^  =  41 

y  =  3x  +  21  •x2  =  2p2/j- 

^-  x^  +  2/2  =  4i"  ^"^-  (0,  0),  (2p,  2j^). 

^rw.  (0,  2),  (- I,  -  f).  9.  ^^'  +  ^'  =  ^1.. 

4.pJ!'o}'  ^«..(J,2),(!.-2,. 

^ns.  (0,  0),  (16,  16).  x^  +  y^  =  100^ 

x2  +  ,/  =  a2        ^  10-  ^2^^              [• 

•  S,  +  y  +  a  =  0j-  2            J 

/.ji^-.   (0,  -a),(-  — ,^).  ^7W.  (8,  6),  (8,  -  6). 


2/2  =  16"!  jj_  a'2+  y2  =  5„2| 


x^  =  »?/ 


x^  =  iay  J 

ylTis.   (±  4  V2,  4).  .•l?is.  (2r(.  a),  (—  2a,  a). 

Find  the  area  of  the  triangles  and  polygons  whose  sides  are  the  loci 
of  the  following  equations  : 

12.  3x  +  y  +  4  =  0,  3x  -  5  2/  +  34  =  0,  3x  -  2?/  +  1  =  0.  Ans.    36. 

13.  X  +  2?/  =  5,  2x  +  ?/  =  7,  ?/  =  X  +  1.  Ans.    |. 

14.  x  +  y  =  «,  X  —  2  ij  =  4  a,  y  —  x  +  7  a  =  0.  Ans.    12  a^. 

15.  X  =  0,  ?/  =  0,  X  =  4,  ?/  =  -  6.  Ans.    24. 

16.  X  —  y  =  0,  X  +  y  =  0.  X  —  ?/  =  a,  X  +  1/  =  6.  ^ns.   -^  ■ 

17.  1/  =  3x  -  9,  ?/  =  3x  +  5,  2 2/  =  X  -  6,  2 2/  =  X  +  14.        ^ns.    56. 

18.  Find  the  distance  between  the  points  of  intersection  of  the  curves 
3x  -  2  ?/  +  6  =  0,  x2  +  v/2  =  9.  Ans.    jf  VlS. 

19.  Does  the  locus  of  ?/2  =  4  x  intersect  the  locus  of2x  +  3r/  +  2  =  0? 

Ans.   Yes. 

20.  For    what    value    of    a    will    the    three    lines    3x+?/  — 2  =  0, 
f'x  +  2  ?/  —  3  =  0,  2  X  —  ?/  —  3  =  0  meet  in  a  i^oint  ?  Aiw.    a  =  5. 

21.  Find    the    length   of   the    common    chord    of    j-'^  +  t/^  =  IS   and 
y^  =  Sx  +  S.  Ans.    (i. 

22.  If  the  equations  of  the  sides  of  a  triangle  are  x  +  7y  +  ll  =  0, 
3x  +  ?/  —  7  =  0,  X  —  3?/  +  l  =  0,  find  the  length  of  each  of  the  medians. 

Ans.    2\/5,  |\/2,  J  VlTO. 


CHAPTER  IV 

THE  STRAIGHT  LIPfE 

V  24.  The  degree  of  the  equation  of  any  straight  line.  It  will 
now  be  shown  that  any  straight  line  is  represented  by  an  equa- 
tion of  the  first  degree  in  the  variable  coordinates  x  and  y. 

Theorem.  The  eqiiation  of  the  straight  line  passing  through  a 
point  B  (0,  l>)  on  the  axis  of  y  and  having  its  slope  equal  to  m  u< 
(I)  y=mx-\-b. 

Proof    Assume  that  P  (x,  ?/)  is  any  point  on  the  line. 
The  given  condition  may  be  written 

slope  of  PB  =  m. 
Since  by  (II),  p.  17,  ^ 

slope  of  PB  =  ^^  ~  ^, 
X  —  0 

[Substituting  {x,  y)  for  (j-^,  y^)  and  (0,  h)  for  (Xg,  y^-l 

1                                  y  —  b  , 

then  ' =  m,  or  y  =  7nx:  -{■  b.  Q.  E.  D. 

X 

In  equation  (I),  m  and  b  may  have  any  values,  positive, 
negative,  or  zero. 

Equation  (I)  will  represent  any  straight  line  which  inter- 
sects the  ?/-axis.  But  the  equation  of  any  line  2)araUel  to  the 
?/-axis  has  the  form  a;  =  a  constant,  since  the  abscissas  of  all 
points  on  such  a  line  are  equal.  The  two  forms,  y  =  mx  -f-  b 
and  X  =  constant,  will  therefore  represent  all  lines.  Each  of 
these  equations  being  of  the  first  degree  in  x  and  y,  we  have  the 

Theorem.  The  equation  of  any  straight  line  is  of  the  first 
degree  in  the  coordinates  x  and  y. 

58 


THE  STRAIGHT  LINE  5& 

r    25.    Locus  of  any  equation  of  the  first  degree.    The  question 
now  arises :    Given   an   equation   of  the  first  degree   in  the 
coordinates  x  and  y,  is  the  locus  a  straight  line  ? 
Consider,  for  example,  the  equation 

(1)  3.r- 2^  +  8  =  0. 

Let  us  solve  this  equation  for  y.    This  gives 

(2)  Z/=t-''  +  ^- 
Comparing  (2)  with  the  formula  (I), 

y  =  vix  +  b, 

we  see  that  (2)  is  obtained  from  (I)  if  we  set  m  =  |,  Z»  =  4. 
Now  in  (I)  m  and  h  may  have  any  values.  The  locus  of  (I)  is, 
for  all  values  of  ??i  and  h,  a  straight  line.  Hence  (2),  or  (1),  is 
the  equation  of  a  straight  line  through  (0,  4)  with  the  slope 
equal  to  |.  This  discussion  prepares  the  way  for  the  general 
theorem. 

The  equation 

(3)  Ax+By  +  C  =  0, 

where  A,  B,  and  C  are  arbitrary  constants,  is  called  the  general 
equation  of  the  first  degree  in  x  and  y  because  every  equation  of 
the  first  degree  may  be  reduced  to  that  form. 
Equation  (3)  represents  all  straight  lines. 

For  the  equation  y  —  mx  +  b  may  be  written  mx  —  y  +  b  =  0,  which 
is  of  the  form  {S)  it  A  =  m,  B  =  —  1,  C  =  b  ;  and  the  e(}uation  x  =  con- 
stant may  be  written  x  —  constant  =  0,  which  is  of  the  form  (3)  if  A  =  1, 
B  =  0,  C  =—  constant. 

Theorem.    The  locns  of  the  general  equation  of  t lie  first  degree 

Ax  -{-By  +  C  =  0 
is  a  straight  line. 

Proof.    Solving  (3)  for  y,  we  obtain 

(4)        ,  y=-T--B 


60 


NEW  ANALYTIC  GEOMETRY 


Comparison   with   (I)   shows  that  the  locus   of  (4)   is   the 

straisfht  line  for  which 

A  C 

')n  =  —  —I  6=— — . 
B  B 

If,  however,  B  =  0,  the  reasoning  fails. 
But  if  5  =  0,  (3)  becomes 

Ax-\-C  =  Q, 

C 
or  X  =  —  —• 

A 

The  locus  of  this  equation  is  a  straight  line  parallel  to  the 

?/-axis.    Hence  in  all  cases  the  locus  of  (3)  is  a  straight  line. 

Q.E.D. 

Corollary.    The  slope  of  the  line 

Ax  +%  +  C  =  0 

Is  VI  =  —  — ;  that  Is,  the  coefficient  of  x  with  its  sign  changed 
B 

divided  hy  the  coefficient  of  y. 

26.  Plotting  straight  lines.  If  the  line  does  not  pass  through 
the  origin  (constant  term  not  zero,  p.  47),  find  the  intercepts 
(p.  47),  mark  them  off  on  the  axes,  and  draw  the  line.  If  the 
line  passes  through  the  origin,  lind  a  second  point  whose 
coordinates  satisfy  the  equation,  and  draw  a  line  through  this 
point  and  tlie  origin. 

EXAMPLE 

Plot  the  locus  of  3  X  —  2/  +  G  =  0.   Find  the  slope. 
Solution.    Letting  ?/  =  0  and  solving  for  x, 

X  =  —  2  =  intercept  on  x-axis. 
Letting       x  =  0  and  solving  for  y, 

2/  =  6  =  intercept  on  ?/-axis. 
The   required   line   passes   through    the    points 
(-2,  0)  and  (0,  6). 

To  find  the  slope :  Comparison  with  the  general 
equation  (3)  shows  that   yl  =  3,  B  =  —  1,  (7  =  6.    Hence  in=.—  —  —  ^. 

Otherwise  thus :  Reduce  the  given  equation  to  the  form  y  =  mx  +  b 
by  solving  it  for  y.   This  gives  ?/  =  3x  +  6.   Hence  m  =  3,  6  =  6,  as  before. 


THE  STRAIGHT  LINE 


61 


-  5,  2i  ;  m  =  i 
-1,  3;  m  =  3. 
i,  3;  m=-|. 


.\C''  PROBLEMS 

1.  Find  the  intercepts  and  the  slope  of  the  following  lines,  and  plot 
the  lines : 

(a)  2  X  +  3  2/  =  6.  Ans.   3,  2  ;  m  =  -  |. 

(b)  x  —  2y+5  =  0.  Ans. 
>'(c)  3x  — ?/  + 3  =  0.  Ans. 
^  (d)  5x  +  2?/  —  6  =  0.  Ans. 

2.  Plot  the  following  lines  and  find  the  slope : 

(a)2x-37/  =  0.  (c)3x  +  22/  =  0. 

(b)  y  -4x  =  0.  {d)  x-3y  =  0. 

K  3.  Find  the  equations,  and  reduce  them  to  the  general  form,  of  the 
lines  for  which 

(a)m  =  2,6=-3.  Ans.    2x  —  y  —  S  =  0. 

(b)  jn=-i,  6=  5.  Ans.   x  +  2y  — 3  =  0. 

^  (c)  m  =  |,6=  —  f.  ^?xs.   4x-10?/- 25  =  0. 

Ans.    X  —  y  —  2  =  0. 


l{d)  a  =  -,b=-2. 

(e)  a  =  —  J>  =  S. 
^  '  4 


Ans.   X  +  y  —  3  =  0. 


Hint.   Substitute  imj  =  mx  +  b  and  transpose. 
^  4.  Select  pairs  of  parallel  and  perpendicular  lines  from  the  following : 
'  L^:y  =  2x-3. 
L^:y=-3x-\-2. 

L,:y  =  2x+  7. 


(a) 


Ans.  L^  II  ig  ;  ^2  -^  -^4- 


^L^:y  =  ix  +  i. 

^^L^:x  +  3y  =  0. 

(b)  -!  ig  :  8x  +  2/  +  1  =  0.  Ans.  L^  ±  ig. 

[Lg:Qx-3y  +  2  =  0. 
fij  :2x-  5?/  =  8. 

(c)  <  L^:  5y  +  2x  =  8.  Ans.  L^  ±  ig. 

[ig:85x-14?/  =  8. 

t^  5.  Show  that  the  quadrilateral  whose  sides  are  2x  —  32/+4  =  0, 
3x  — ?/  — 2  =  0,  4x  — 6^  — 9  =  0,  and  6x  —  2y  +  4  =  0  is  a  paral- 
lelogram. 

6.  Find  the  equation  of  the  line  whose  slope  is  —  2,  which  passes  through 
the  point  of  intersection  of  y  =  3x  +  4  and  y  =—  x  +  4. 

Ans.    2x  +  y  —  4  =  0. 


62  NEW  ANALYTIC   GEOMETRY 

^     7.  Write  an  equation  which  will  represent  all  lines  parallel  to  the  line 
(a)  2/  =  2a; +7.  (c)  y  _  3x  -  4  =  0. 

{b);/=-a;  +  9.  (d)  2?/ -  4x  +  3  =  0. 

8.  Find  the  equation  of   the   line   parallel  to   2x  —  ^y  =  Q   whose 
intercept  on  the  F-axis  is  —  2.  Ans.    2x  —  3j/  —  6  =  0. 

9.  Show  that  the  following  loci  are  straight  lines  and  plot  them  : 

(a)  The  locus  of  a  point  whose  distances  from  the  axes  XX'  and  YY' 
are  in  a  constant  ratio  equal  to  |.  Ans.    2x  —  3?/  =  0. 

(b)  The  locus  of  a  point  the  sum  of  whose  distances  from  the  axes  of 
coordinates  is  always  equal  to  10.  Ans.   a;  +  ?/  —  10  =  0. 

(c)  A  point  moves  so  as  to  be  always  equidistant  from  the  axes  of 
coordinates.  Ans.   x  —  y  =  0. 

(d)  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  (3,  0)  and  (0,  —  2)  is  always  equal  to  8. 

Ans.    The  parallel  straight  lines  6x  +  42/  +  3=:0,  6x  +  42/  —  13  =  0. 

(e)  A  point  moves  so  as  to  be  always  equidistant  from  the  straight 
lines  X  —  4  =  0  and  2/  +  5  =  0. 

Ans.   The  perpendicular  straight  lines  x  —  y  —  9  =  0,  x  +  y  +  1—  0. 

10.  A  point  moves  so  that  the  sum  of  its  distances  from  two  perpen- 
dicular lines  is  constant.    Show  that  the  locus  is  a  straight  line. 

Hitit.  Choosing  the  axes  of  coordinates  to  coincide  with  the  given  lines,  the 
equation  \sx  +  y  —  constant. 

11.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  two  fixed  points  is  constant.  Show  that  the  locus  is  a  pair  of 
straight  lines. 

Hint.  Draw  A'A''  through  the  fixed  points,  and  YY'  through  their  middle 
point.  Then  the  fixed  points  may  be  written  (a,0),  (-  a,0),  and  if  the"  constant 
difference  "  be  denoted  by  k,  we  find  for  the  locus  4  ax  =  k  and  4  ax  =-  k. 

12.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  two  perpendicular  lines  is  zero.  Show  that  the  locus  is  a  pair  of 
perpendicular  lines. 

13.  A  point  moves  so  that  its  distance  from  a  fixed  line  is  in  a  constant 
ratio  to  its  distance  from  a  fixed  point  on  the  line.  For  what  values  of  the 
ratio  is  the  locus  real  ?  What  is  the  locus  ? 

27.  Point-slope  form.  If  it  is  required  that  a  straight  line 
shall  pass  through  a  given  point  in  a  given  direction,  the  line 
is  determined. 


THE  STRAIGHT  LINE  63 

The  following  problem  is  therefore  definite : 

To  find  the  equation  of  the  straight  line  passing  throxigh  a 
given  -point  P^(x^,  y,)  and  having  a  given  slope  m. 

Solution.  Let  P(x,  y)  be  any  other  point  on  the  line.  By 
the  hypothesis,  ^^^^  pp^  ^  ^^ 

(1)  •'•^^^  =  ^^-  (n,p.i7) 

Clearing  of  fractions  gives  the  formula 

(II)  y-yi  =  m{x-x^). 

28.  Two-point  form.  A  straight  line  is  determined  by  two 
of  its  points.    Let  us  then  solve  the  problem : 

To  find  the  equation  of  the  line  passing  through  two  given 
points  P^(x^,  ?/j),  P^(x^,  y.^. 

Solution.    The  slope  of  the  given  line  is 

slopeP,P,  =  ^5^^- 
Let  P  (x,  y)  be  any  other  point  on  the  line  P^P^-    Then 

slope  pp, = y  ~y^ . 

^      a*  —  Xj 

Since  P,  Pj,  and  P„  are  on  one  line,  slope  PP^  =  slope  PJ*^- 
Hence  we  have  the  formula 

(III)  y-Vx    Vx-Vi, 


M 


X^    — —  Xn 


Equation  (III)  may  be  written  in  the  determinant  form 
X      y      \ 
(2)  Xi     z/i     1    =0. 


X 

y     1 

Xi 

Vi    1 

^1 

2/2      1 

For  the  determinant,  when  expanded,  is  of  the  first  degree  in  x  and  y. 
Hence  (2)  is  the  equation  of  a  line.  But  (2)  is  satisfied  when  x  =  x^, 
y  =  ?/j,  and  also  when  x  =  x^,  y  =  yn,  for  then  two  rows  become  identical 
and  the  determinant  vanishes.  Otherwise  thus  :  Comparison  of  (2)  with 
the  formula  at  the  close  of  Art.  14  shows  that  the  area  of  the  triangle 
PP^P,,  is  zero.    Hence  these  three  points  lie  on  a  line. 


64 


NEW  ANALYTIC  GEOMETRY 


3, 


EXAMPLES 

1.  Find  the  equation  of  the  line  passing  througli  P^  (3,  —  2)  whose 
slope  is  —  J. 

Solution.     Use  the  point-slope 
equation  (II),  substituting  x^  = 
?/j  =  —  2,  m  =—  I.    This  gives 

y  +  2=-^{X-S). 

Clearing  and  reducing, 
X  +  4y  +  5  =  0. 

2.  Find  the  equation  of  the  line  through  the  two  points  P^  (5,  —  1)  and 
^2(2,-2). 

Solution.    Use  the  two-point  equation  (III),  substituting  Xj  =  5,  ?/j  =  —  1, 


P  ( 3,-2> 


= 

2, 

2/2  =-2 

This  gives 

y  +  i 

-1  +  2 

1 

x-5 

5-2 

3 

CI 

earing  and 

reducing. 

X  — 

Sy-8^0. 

The  answer  should  be  checked.  To  do  this,  we  must  prove  that  the  coor- 
dinates of  the  given  points  satisfy  the  answer.  Thus  for  Pj,  substituting 
X  =  5,  2/  =—  1,  the  answer  holds.  Similarly  for  P^.  The  student  should 
supply  checks  for  Examples  1,  3,  and  4. 

3.  Find  the  equation  of  the  line  through 
the  point  P^  (3,  —  2)  parallel  to  the  line 
L^:2x-Sy-i  =  0. 

Solution.  The  slope  of  the  given  line 
ij  equals  |.  Hence  the  slope  of  the  re- 
quired line  also  equals  |  (Theorem,  p.  17), 
and  it  passes  through  Pj(3,  —  2).  Using 
the  point-slope  equation  (II),  we  have  y  +  2 
=  1  (a;-3),  or  2X-32/-  12  =  0. 

4.  Find  the  equation  of  the  line  through  the  point  Pi(—  1,  3)  perpen- 
dicular to  the  line  i^:5x  —  22/  +  3  =  0. 

Solution.  The  slope  of  the  given  line  L^ 
equals  |.  Hence  the  slope  of  the  required 
line  equals  —  |  (Theorem,  p.  17).  Since  we 
know  a  point  Pi(—  1,  3)  on  the  line,  we 
use  the  point-slope  equation  (II),  and  obtain 
y  -  3  =  -  Kx  +  1),  or  2  X  4-  5  ?/  -  13  =:  0. 


THE  STRAIGHT  LINE  65 

PROBLEMS 

1.  Find  the  equation  of  the  line  satisfying  the  following  conditions, 
and  plot  the  line.    Check  the  answers  : 

(a)  Passing  through  (0,  0)  and  (8,  2).  Ans.   x  —  4y  =  0. 

(b)  Passing  through  (—  1,  1)  and  (—  3,  1).  Ans.    tj  —  1  =  0. 

(c)  Passing  through  (—  3,  1)  and  slope  =  2.        Ans.    2x  —  y  +  7  =  0. 

(d)  Having  the  intercepts*  a  =  3  and  6  =—  2.     Ans.  2x  —  Sy—  6  =  0. 

(e)  Slope  =—3,  intercept  on  cc-axis  =  4.  Ans.   3x  +  v/  —  12  =  0. 

(f)  Intercepts  a  =—3  and  6  =—  4.  Ans.   Ax  +  Sy  +  12  =  0. 

(g)  Passing  through  (2,  3)  and  (—  2,  -  3).  Ans.  3x  —  2y  =  0. 
(h)  Passing  through  (3,  4)  and  (—  4,  —  3).  Ans.  x  —  y  +  1  -  0. 
(1)  Passing  through  (2,  3)  and  slope  =—2.  Ans.    2x  +  y  —  7  =  0. 

2.  Find  the  equation  of  the  line  passing  through  the  origin  parallel 
to  the  line  2a;  —  3?/ =  4.  Ans.    2x  —  3y  =  0. 

3.  Find  the  equation  of  the  line  passing  through  the  origin  perpen- 
dicular to  the  line  5x  +  y  —  2  —  0.  Ans.   x  —  ^y  =  Q. 

4.  Find  the  equation  of  the  line  passing  through  the  point  (3,  2)  par- 
allel to  the  line  4x  —  ?/  —  3  =  0.  Ans.    4x  —  y  —  10  =  0. 

5.  Find  the  equation  of  the  line  passing  through  the  point  (3,0)  per- 
pendicular to  the  line  2x  +  y  —  o  =  0.  Ans.   x  —  2  y  —  3  =  0. 

6.  Find  the  equation  of  the  line  whose  intercept  on  the  ?/-axis  is  5, 
which  passes  through  the  point  (6,  3).  Ans.   x  +  Sy  —  15  =  0. 

7.  Find  the  equation  of  the  line  whose  intercept  on  the  a; -axis  is  3, 
which  is  parallel  to  the  line  x  —  Ay  +  2  =  0.  Ans.   x  —  4?/  —  3  =  0. 

8.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  —  2y  +  S  =  0  and  x  +  2y  —  9  =  0. 

Ans.   x  —  y  =  0. 

9.  Find  the  equations  of  the  sides  of  the  triangle  whose  vertices  are 
(-  3,2),  (3,  -2),  and(0,  -1). 

Ans.    2x-l-32/  =  0,  x-f-32/-l-3  =  0,  and  x  +  y  +  1  =0. 
10.  Find  the  equations  of  the  medians  of  the  triangle  in  Problem  9,  and 
show  that  they  meet  in  a  point. 

Ans.    X  =  0,  7 X  +  9 y  +  3  =  0,  and  fjx  +  dy  +  3  =  0. 

Hint.    To  show  that  three  lines  meet  in  a  point,  tind  the  point  of  intersection 
of  two  of  them  and  prove  that  it  lies  on  the  third. 

*  Intercept  on  avaxis  =  a,  intercept  on  y-axis  =  b.  The  given  points  are  (3,  0) 
and  (0,  -  2) . 


66  NEW  ANALYTIC  GEOMETRY 

11.  Determine  whether  or  not  the  following  sets  of  points  lie  on  a 

straight  line  : 

(a)  (0,  0),  (1,  1),  (7,  7).  Ana.   Yes. 

(b)  (2,  3),  (-  4,  -  6),  (8, 12).  Ans.  Yes. 

(c)  (3,4),  (1,2),  (.5,1).  Ans.  No. 

(d)  (3,  -  1),  (-  6,  2),  (-  I  1).  Ans.  No. 

(e)  (5,6),  (i,l),  (-1,-5).  Ans.  Yes. 

(f )  (7,  6),  (2,  1),  (0,  -  2).  Ans.  No. 

(g)  (3,  -  2),  (6,  -  4),  (-  5,  4). 
(h)  (1,  0),  (0,  1),  (7,  -  8). 

(i)  (-3,  -1),  (6,  2),  (8,3). 

12.  Find  the  equations  of  the  lines  joining  the  middle  points  of  the  sides 
of  the  triangle  in  Problem  9,  and  show  that  they  are  parallel  to  the  sides. 

Ans.    4x  +  6y+  3  =  0,  j  +  3y  =  0,  and  x  +  y  =  0. 

13.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  +  2  y  =  1  and  2x  —  -iy  —  S  =  0. 

Ans.   X  +  10 y  -0. 

14.  Show  that  the  diagonals  of  a  square  are  perpendicular. 
Hint.   Take  two  sides  for  the  axes  and  let  the  length  of  a  side  be  a. 

15.  Show  that  the  line  joining  the  middle  points  of  two  sides  of  a  tri- 
angle is  parallel  to  the  thiixl. 

Hint.   Choose  the  axes  so  that  the  vertices  are  (0,0),  (</,  0),  and  (b,  c). 

16.  Two  sides  of  a  parallelogram  are  given  hj2x  +  Sy— 7  =  0  and 
X  —  3y  +  4  =  0.    Find  the  other  two  sides  if  one  vertex  is  the  point  (3,  2). 

Ans.    2x  +  Sy-12  =  0  and  x  -  3  //  +  3  =  0. 

17.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  whose  vertices  are  (—  3,  2),  (3,  —  2),  and  (0,—  1),  which  are  par- 
allel to  the  opposite  sides.    Find  the  vertices  of  the  new  triangle. 

Ans.    2x  +  3y  +  3-=0,x  +  3y-3  =  0,x  +  y-l=:0. 

18.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  in  Problem  17,  which  are  perpendicular  to  the  opposite  sides,  and 
show  that  they  meet  in  a  point. 

Ans.   3x-2y-2  =  0,8x-y+n  =  0,x-y-5  =  0. 

19.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of 
the  triangle  in  Problem  17,  and  show  that  they  meet  in  a  point. 

^n*.   3x-2y  =  0,Sx  —  y-6  =  0,  X-  y  +  2  =  0. 


THE  STRAIGHT  LINE  67 

20.  The  equations  of  two  sides  of  a  parallelogram  are  3x  —  4i/4-6  =  0 
and  I  +  5i/  —  10  =  0.  Find  the  equations  of  the  other  two  sides  if  one 
vertex  is  the  point  (4,  9).       Aiis.    3x  —  4?/ +  24  =  0  and  x  +  5y  —  49  =  0. 

21.  The  vertices  of  a  triangle  are  (2,  1),  (—  2,  3),  and  (4,  —  1).  Find 
the  equations  of  (a)  the  sides  of  the  triangle,  (b)  the  perpendicular  bisec- 
tors of  the  sides,  and  (c)  the  lines  drawn  through  the  vertices  perpendicu- 
lar to  the  opposite  sides.  Check  the  results  by  showing  that  the  lines  in 
(b)  and  (c)  meet  in  a  point. 

29.  Intercept  form.  A  line  is  determined  if  its  intercepts  on 
the  axes  are  given.  If  these  intercepts  are  a  on  A' A''  and  b  on 
YY',  then  the  line  passes  through  (a,  0)  and  (0,  b),  and  the  two- 
pcint  form  (III)  gives  (writing  x^  —  a,  y^  =  0,  x^  =  0,  y,^  =  h) 

X  —  a       0  —  rt  a 

Clearing  of  fractions,  transposing,  and  dividing  by  ah,  we 
obtain 

(IV)  ^  +  ^  =  1. 

30.  Condition  that  three  lines  shall  intersect  in  a  common 
point.  It  is  shown  in  algebra  that  three  linear  equations  in 
two  unknoAvns  x  and  y,  for  example, 

(1)    Ax  +  ny+C  =  0,    A^x  +  l\y-\-C\^i),    A.^x  +  B^y-{- C^=Q, 

will  have  a  common  solution  when  and  only  when  the  deter- 
minant formed  on  the  coefficients  vanishes ;  that  is,  when 


(2) 


A 

B 

C 

Ay 

1\ 

(\ 

A„ 

L'„ 

C, 

0. 


Hence  the  three  lines  (1)  will  intersect  in  a  common  point 
when  and  only  when  (2)  holds,  provided  always  that  the  lines 
are  not  jtarallel,  however.  But  this  latter  fact  may  always  be 
determined  by  inspection  of  the  e(]uations. 


68 


NEW  ANALYTIC   GEOMETRY 


--^P 


if 


X' 


MX 


W 


31.  Theorems  on  projection.  In  preparation  for  deriving  addi- 
tional theorems  of  this  and  later  chapters,  some  simple  facts 
in  regard  to  projection  will  now  be  discussed. 

The  orthogonal  projection  of  a  point  upon  a  line  is  the  foot  of 
the  perpendicular  let  fall  from  the  point  upon  the  line. 

Thus  in  the  figure 

M  is  the  orthogonal  projection  of  P 

on  A' 'A'; 
N  is  the  orthogonal  projection  of  P 

on  Y'Y-. 
P'  is  the  orthogonal  projection  of  P' 

on  A" A. 

If  A  and  B  are  two  points  of  a  directed  line,  and  ^f  and  N 
their  projections  upon  a  second  directed  line  CD,  then  MN  is 
called  the  projection  of  AB  upon  CD. 

First  Theorem  of  Projection.  If  A  and  B  arejjoints  upon 
a  directed  line  making  an,  angle  a  with  a  second,  directed  line 
CD,  then  the 

projection  of  the  length  AB  upon  CD  =  AB  cos  a. 
Proof.    In  the  iigures 

projection  of  AB  upon  CD  =  MN. 

B  B 


r1 


oy 


N  D 


C    N 


Fig.  1  Fig.  2 

Now  in  Fig.  1,  from  the  right  triangle  BAS, 

AS  =  AB  cos  BAS. 
But  A  S  =  MN,  and  Z  BA  S  =  a. 

.•.  MN  =  AB  cos  a. 


THE  STRAIGHT  LINE  69 

In  Fig  2  (p.  68),  a  is  obtuse  and  MN  is  a  negative  number. 
Numerically,  AS  and  MN  are  equal,  but  they  differ  in  sign, 
A S  not  being  directed.    As  before,  AS  =  AB  cos  BA S.    But 
ZBAS=  180°  -a.     .-.  cos  BAS  =  -  cos  a  (30,  p.  3). 
Hence  AS  =  —  AB  cos  a. 

.'.  AIX  =  AB  cos  a.  Q. E. D. 

Consider  next  a  broken  line  made  up  of  directed  parts,  as  in 
the  figures.  The  line  joining  the  first  and  last  points  of  a 
broken  line  is  called  the  closing  line. 

B> 
Pi 

Mi  D 

Fia.  1 

Thus  in  Fig.  1  the  closing  line  is  P^P^ ;  in  Fig.  2  the  closing 
line  is  P^P^- 

With  reference  to  such  broken  lines,  the  following  theorem, 
of  frequent  application,  holds. 

Second  Theorem  of  PROJECxioisr.  If  each  segment  of  a 
hroTien  line  he  given  the  direction  determined  injxissing  continu- 
oti sly  from  one  extremity  to  the  other,  then  the  algebraic  sum  of 
the  2}roJectio?is  of  the  segments  ujion  any  directed  line  equals 
the  projection  of  the  closing  line. 

Proof    The  proof  results  immediately.    For,  in  Fig.  1, 

M^il/^  =  projection  of  Pf'.-^; 

M^M^  —  projection  of  P.f'^ ; 

M^M^  =  projection  of  closing  line  P-^P^. 
But  obviously       M^M^  +  MJVI^  =  M^M^, 
and  the  theorem  follows. 

Similarly  in  Fig.  2.  Q.  e.  d. 


70 


NEW  ANALYTIC   GEOMETRY 


Corollary.  If  the  sides  of  a  closed  polygon  he  given  the  direc- 
tion established  by  passing  contimtously  around  the  perimeter,  the 
sum  of  the  p/rojections  of  the  sides  upon  any  directed  line  is  zero. 

For  the  closing  line  is  now  zero. 

32.  The  normal  equation  of  the  straight  line.  In  the  preceding 
sections  the  lines  considered  were  determined  by  two  points  or 
by  a  point  and  a  direction.  Both  of  these  methods  of  determin- 
ing a  line  are  frequently  used  in  elementary  geometry,  but  we 
have  now  to  consider  a  line  determined  by  two  conditions  which 
belong  essentially  to  analytic  geometry.  Let  AB  be  any  line, 
and  let  ON  be  drawn  from 
the  origin  perpendicular  to 
Ali  at  C.  Let  the  j^ositive 
divcrtion  on  ON  be  from  O 
toward  N,  that  is,  from  the 
origin  toward  the  line,  and 
denote  the  positive  directed 
length  OC  by /i, and  the  posi- 
tive angle  A'OiV,  measured, 
as  in  trigonometry,  from  OX 
as  initial  line  to  ON  as  ter- 
minal line,tjf  o) (Greek  letter  "omega").  Then  it  is  evident  from 
the  figures  that  the  position  of  any  line  is  deterTnined  by  a  jniir 
of  values  ofp  and  w,  both  j'  and  to  being  pjositive  and  w  <  3C0°. 

On  the  other  hand,  every  line  which  does  not  pass  through 
the  origin  determines  a  single  positive  value  of  ^  and  a  single 
positive  value  of  w  which  is  less  than  360°. 

The  problem  now  is  this  :  Given  for  the  line  AB  of  the  figure 
the  perpendicular  distance  OC  (=7>)  from  the  origin  and  the 
angle  XOC  (=  w);  to  find  the  equation  of  AB. 

Solution.    Let  P(x,y)  be  any  point  on  the  given  line  AB. 
Then  since  yl/>  is  perpendicular  to  ON,  the  projection  of 
OP  on  ON  is  equal  to  pj.    Consider  the  broken  line  ODP.    The 


THE  STRAIGHT  LINE 


71 


closing  line  is  OP.  By  the  second  theorem  of  projection  (p.  69), 
the  projection  of  OP  on  OX  is  equal  to  the  sum  of  the  projec- 
tions of  OD  and  DP  on  ON.    Then 

(1)  projection  of  OD  on  OX  +  projection  of  DP  on  ON  =^p. 

By  the  first  theorem  of  projec- 
tion (p.  68), 

(2)  projection  of  OD  on 
ON  =  OD  cos  iii  =x  cos  o),  and 

(3)  projection  of  DP  on 


OX  =  DP  coslZ  - 


<f-») 


1/  sni  o). 


[For  the  angle  between  the  directed  lines  DP  and   ON  equals  that 
between  OY  and  ON  = w.] 


Substituting  from  (2)  and  (3)  in  (1), 
(V)  x  cos  CO  -|-  1/  sin  CO  —  />  =  0. 


Q.  E.D. 


.v\ 

\ 

s^ 

z^-- 

^x* 

^ 

X 

'^ 

\x' 

This  equation  is  known  as  the  normal  eq nation. 

^Yhen^;  =  0,  however,  AB  passes  through  the  origin,  and  the 
rule  given  above  for  the  posi- 
tive direction  on  OX  becomes 
meaningless.  From  the  fig- 
ures we  see  that  we  can 
choose  for  w  either  of  the 
an  gles  A'  OX  or  A'  6* A' '.  IVh  en 
J/  =z  0  ive  sJuill  alir<ii/s  suppose  that  w  <C  180°  and  that  the  p>osl- 
tlve  (Jlrcctum.  on  <>N  is  the  iijucard  directum. 

33.  Reduction  to  the  normal  form.  In  Art.  25  it  appeared  that 
the  slope  of  any  line  could  be  found  after  its  equation  was 
reduced  to  the  form  y  =  mx  +  h.  If  now  the  equation  of  any 
line  can  be  reduced  to  the  normal  form  (V),  we  shall  be  able  to 
find  the  perpendicular  distance  p  from  the  origin  to  the  line, 
as  well  as  the  ingle  w  which  this  perpendicular  makes  with  OX. 


72  NEW  ANALYTIC  GEOMETRY 

To  reduce  a  given  equation 

(1)  Ax  +  lU/  +  C  =  0 

to  the  normal  form,  it  is  necessary  to  determine  w  and^  so  that 
the  locus  of  (1)  is  identical  with  the  locus  of 

(2)  X  cos  w  +  y  s^i^  0)  —  2^  =  0. 

This  is  the  case  when  corresponding  coefficients  are  propor- 
tional.*   Hence  we  must  have 

cos  w  _  sin  <a  _—  p 
A     ~     B     ~  ~c" 

Denote  the  common  value  of  these  ratios  by  r ;  then 

(3)  cos  w  =  vA , 

(4)  sin  oj  =  vB,  and 

(5)  —  P  =  ^'C 

To  find  r,  square  (3)  and  (4)  and  add ;  this  gives 

sin^o)  +  cos^o)  =  7^{A^  +  B^). 
But  sin^w  +  cos'^o)  =  1 ;  (28,  p.  3) 

and  hence  r^(^A'^  -\-  B^)  =  1,  or 

(6)  .  r  = /  . 

^  ^  ±^A-'  +  B-' 

Equation  (5)  shows  which  sign  of  the  radical  to  use ;  for 
since  jj  is  positive,  r  and  C  must  have  opposite  signs. 
Substituting  the  value  of  r  in  (3),  (4),  and  (5), 

A  .  B  C 

;    Sin  o)  = .  )  p  = ■ 

±  V^2  +  B'  ±  WA'^  -h  B'^  ±  WA^  -j-  B' 


cos  0)  = 


Hence  (2)  becomes 

(0       ,  X  H ,  y  -\ ,  =  0, 

^^        ^^A-'  +  B'        ±^A''  +  B^        ±  V/l-^  +  B' 

*  The  proof  of  this  obvious  fact  is  left  to  the  sti  dent. 


THE  STRAIGHT  LIXE  73 

which  is  the  normal  form  of  (1).  The  result  of  the  discussion 
may  be  stated  in  the  following 

Rule  to  reduce  Ax  +  J3y  -\-  C  ^  0  to  the  normal  form. 

Find  the  numerical  value  of  V.l'^  +  -^^  «^<^  give  it  the  sign 
opposite  to  that  of  C.  Divide  the  given  equation  hy  this  number. 
The  result  is  the  required  equation. 

For  example,  to  reduce  the  equation 
(8)  3x-?/  +  10  =  0 

to  the  normal  form,  divide  the  equation  by  —  VlO,  since  -4  =  3,  U  =  —  1, 
^A-  -\-  B^  =  VlO ,  and  this  radical  must  be  given  the  negative  sign, 
since  C  {=  10)  is  positive.    The  normal  form  of  (8)  is  accordingly 

=  X  H =  y  —  v'lO  =  0. 

VlO       VlO 

Here  casw  = =,  sin  w  =  — =5  p  —  VlO  =  3.1  +. 

VlO  VlO 

If  C  =  0,  then  ^  =  0,  and  hence  a>  <  180°  (p.  71)  ;  then 
sin  (ri  is  positive,  and  from  (4)  r  and  B  must  have  the  same  signs. 

The  advantages  of  the  normal  form  of  the  equation  of  the 
straight  line  over  the  other  forms  are  twofold.  In  the  first 
place,  every  line  may  have  its  equation  put  in  the  normal  form  ; 
whether  it  is  parallel  to  one  of  the  axes  or  passes  through  the 
origin  is  immaterial.  In  the  second  place,  as  will  be  seen  in  the 
following  section,  it  enables  us  to  find  immediately  the  perpen- 
dicular distance  from  a  line  to  a  point. 

PROBLEMS 

1.  In  what  quadrant  will  ON  (see  figure  on  page  70)  lie  if  sin  w  and 
cos  w  are  both  positive  ?  both  negative  ?  if  sin  w  is  positive  and  cos  a> 
negative  ?  if  sin  w  is  negative  and  cos  oj  positive  ? 

2.  Find  the  equations  and  plot  the  lines  for  which 

•     (a)  w  =  0,  p  =  5.  An&.    x  =  6. 

(b)a;  =  ^,p  =  3.  Ann.    ^  +  3  =  0. 

\.(c)  o)  = -,p  =  3.  Am.    \/2x  +  V27/ -  6  =  0. 

4 


74  NEW  ANALYTIC  GEOMETRY 

2  ir  r- 

v^/  (d)  w  =  —  ,  P  =  2.  Ans.    X  —  V3  2/  +  4  =  0. 

O 

(e)  oj  =  ■^,p  =  4.  ^TW.    V2x  — V2i/  — 8  =  0. 

4 

3.  Reduce  the  following  equations  to  the  normal  form  and  find  p  and  w : 

'(a)  3x  +  4(/  —  2  =  0.  ^ns.   p  =  f,  w  =  cos-i  \  =  sin-i|. 

(b)  3x  -  4?/  —  2  =  0.  ^ns.  p  =  |,  w  =  cos-i  |  =  sin-i  (—  |). 

,  (c)  12x  — 52/  =  0.  -4ns.  p  =  0,  w  =  cos-^  (— 1|)=  sin-i  j^^. 

(d)  2  X  +  5  2/  +  7  =  0. 

^7W.  p  = ^- — ,  u>  =  cos~  1  / — - — )  =  sin-i 


+  V29  \-V29/  \-\''29 

(e)  4a;  —  3 //  +  1  =  0.  A-m.  p  =  I,  u  =  cos-i(—  |)  =  sin-i  | 

(f)  4x-  5y  +  6  =  0. 

Ans.  p 


+  V41  \-V41/  \+^'41 

(g)x-4  =  0.         (h)2/-3  =  0.         (i)x  +  2  =  0.         (j)2/  +  4  =  0. 

4.  Find  the  perpendicular  distance  from  the  origin  to  each  of  the 
following  lines : 

(a)  12x+ 52/-26  =  0.  Ans.    2. 

(b)  x  +  y  +  1  =  0.  Ans.   i  V2. 

(c)  3x-2?/-l  =  0.  Ans.    ^"3  Vl3. 

(d)  X  +  4  =  0. 

(e)  y-b  =  0. 

5.  Derive  (V)  when  (a)  -  <  w  <ir  ;  (b)  tt  <  w  <  - —  ;  (c)  —  <  w  <  2  ir ; 
{6.)p  =  0,  and  0  <  u>  <  -  . 

\       6.  For  what  values  of  p  and  w  will  the  locus  of  (V)  be  parallel  to  the 
X-axis  ?  the  y-axis  ?  pass  through  the  origin  ? 

7.  Find  the  equations  of  the  lines  whose  slopes  equal  —  2,  which  are 
at  a  distance  of  5  from  the  origin. 

Ans.    2V5x  +  v'Sy  —  25  =  0  and  2\/5x  +  VS?/  +  25  =  0. 

8.  Find  the  lines  whose  distance  from  the  origin  is  10,  which  pass 
through  the  point  (5,  10).  --Ins.    y  =  10  and  4x  +  3?/  =  50. 

9.  Write  an  equation  representing  all  lines  whose  perpendicular  dis- 
tance from  the  origin  is  5. 

34.  The  perpendicular  distance  from  a  line  to  a  point.     The 

positive  direction  on  the  line   OX  drawn  through  the  origin 
perpendicular  to  AB  is  from  O  to  AB  (Art.  32).    The  positive 


THE  STRAIGHT  LINE 


75 


direction  on  ON  will  now  be  assumed  to  determine  the  positive 
direction  on  all  lines  perpendicular  to  AB.  Hence  the  perpen- 
dicular distance  from  the  line  AB  to  the  point  P^  is  positive 
if  Pj  and  the  origin  are  on  opposite  sides  of  AB,  and  negative 
if  /'j  and  the  origin  are  on  the  same  side 
(if  AB.  Thus  in  the  figure  the  distance 
from  AB  to  P^  is  positive,  and  from  AB 
to  1\^  is  negative. 

Let  us  now  solve  this  problem :  Given 
the  equation  of  any  line  AB  and  a  point 
P^ ;  to  find  the  perpendicular  distance  from 
A  B  to  /\. 

Solution.    Assume  that  the  equation  of  AB  is  in  the  normal 
form 

(1)    X  cos  oi  +  y   sin  w 
-^.  =  0. 

Let  the  coordinates  of 
7'j  be  (x^,  y^  and  denote 
the  perpendicular  dis- 
tance from  A  B  to  P^  by  d. 
In  the  figure  project  the 
broken  line  01)]^^  upon 
the  normal  ON.  Then 
since  OP^  is  the  closing 
line,  by  the  second  theorem  of  projection  (p.  69),  projection  of 
OPj  on  0N=  projection  of  OD  on  OiV+ projection  of  DP^  on  ON. 

From  the  iigure, 

projection  of  OP^  on  ON  =  OE  =p  +  d. 

By  the  first  theorem  of  projection  (p.  GS), 

projection  of  ()]>  on  ON  =  OD  cos  w  =  x^  cos  <o, 
projection  of  DP^  on  ON  =  DP^  cos  i-  —  u>\ 
=  7/j  sin  <D. 


76 


NEW  ANALYTIC  GEOMETRY 


Hence  p  -{-  d  =  x^  cos  w  +  y^  sin  w, 

and  therefore  d  ^  x^  cos  w  +  y^  sin  w  —  js. 


Q.  E.  D. 


In  words  :  The  perpendicular  distance  d  is  the  result  oL- 
tained  by  substituting  the  coordinates  of  P^  for  x  and  y  in  the 
left-hand  member  of  the  normal  equation  (1). 

Hence  the 

Rule  to  find  the  perpendicular  distance  d  from  a  given  line  to 
a  given  point. 

Reduce  the  eqxiation  of  the  given  line  to  the  normal  form 
(Art.  S3),  place  d  equal  to  the  left-hand  member  of  this  equa- 
tion, and  then  suhstitute  the  coordinates  of  the  giiwn  j^oint  for  x 
and  y.    The  residt  is  the  required  distance. 

The  sign  of  the  result  will  show  if  the  origin  and  the  given 
point  are  on  the  same  side  {d  is  negative)  or  opposite  sides 
(d  is  positive)  of  the  line. 

The  perpendicular  distance  d  from  the 
line  Ax  +  By  +  C  =  0  to  the  point  (a-j,  y^) 
will  be,  by  this  rule,  equal  to 

Ax,+By,  +  r 


(2) 


d  = 


±  ^A-  +  B' 


the  sign  of  the  radical  being  opposite  to 
the  sign  of  C. 

When  the  given  line  AB  passes  through  the  origin,  the  posi- 
tive direction  on  the  normal  ON  is  the  upward  direction. 
Hence  the  rule  just  stated  will  give  a  j^ositive  result  for  d  when 
the  perpendicular  drawn  from  the  line  to  the  point  has  the 
upward  direction,  and  a  negative  result  in  the  contrary  case. 
Thus  in  the  figure  the  distance  to  P^  is  positive  and  to  P^  is 
negative. 

Formula  (2)  may  be  used  to  find  the  perpendicular  distance, 
but  it  is  recommended  that  the  rule  be  applied  instead. 


THE  STRAIGHT  LINE 


77 


EXAMPLES 

1,  Find  the  perpendicular  distance  from  the  line  4ic  —  3j/  +  15  =  0to 
the  point  (2,  1). 

Solution.  The  equation  is  reduced  to  the 
normal  form  by  dividing  by  —  Vl6  +  9  =— 5. 
Placing  d  equal  to  the,  left-hand  member  thus 
obtained, 

3?/ +  15 


d  = 


4x 


Substituting  x  =  2,  y  =  1,  then  d  = 


8-3  +  15 


=  -4. 


Hence  the  length  of  the  perpendicular  distance  is 

4  and  the  point  is  on  the  same  side  of  the  line  as  the  origin. 

2.  Find  the  equations  of  the  bisectors  of  the  angles  formed  by  the  lines 
Lj^:x  +  3y-6  =  0, 
L^:Sx  +  y  +  2  =  0. 
Solution.    Let  P^  (Xj,  y^)  be  any  point  on  the  bisector  ig.    Then,  by 
geometry,  P^  is  equally  distant  from  the  given  lines.    Thus,  if 

d^  =  distance  from  L^  to  P^, 
and  dg  =  distance  from  L.^  to  P^, 

then  d■^  and  d.^  are  numerically  equal.    Since,  however,  Pj^  is  on  the  same 
side  of  both  lines  as  the  origin,  dj 
and  dg  are  both  negative.    Hence 
for  every  point  on  the  bisector  ig, 

(1)  d,  =  d,. 

By  the  rule  for  finding  d, 

x,  +  Sy,-6 
"l 7= ' 

Vio 

_3x,  +  y,  +  2 
"2- 7= 

-Vio 

Substituting  in  (1)  and  reducing, 

(2)  Xi  + 2/1-1  =  0. 
Dropping  the  subscripts  in  or- 
der to  follovF  the  usual  custom  of 

having  (x,  y)  denote  any  point  on  the  line,  we  have  for  the  equation  of 

(3)  Xg  :  X  +  y  —  1  =  0.    Ans. 


78  NEW  ANALYTIC   GEOMETRY 

For  any  point  on  the  bisector  L^  the  distances  dj  and  d^  will  be  equal 
numerically  but  will  differ  in  sign.    Hence,  along  L^, 

(4)  di=-d2- 
Proceeding  as  before,  the  equation  of  L^  is  found  to  be 

(5)  L^:x-y  +  4  =  0.    Ans. 

We  note  that  (3)  and  (5)  represent  pei*pendicular  lines. 

Regarded  as  a  formal  process,  equations  (3)  and  (5)  of  the  bisectors 
are  found  by  reducing  the  equations  of  L^  and  L.^  to  the  normal  form  and 
then  adding  and  subtracting  these  efjuations. 

PROBLEMS 

1.  Find  the  perpendicular  distance  from  the  line 

(a)  xcos45°  + 2/sin45=-V2  =  0  to(5,  —  7).  Ans.  — 2V2. 

(b)  I  X  -  1 2/  -  1  =  0  to  (2, 1).  Ans.  -  §. 

(c)  3x  +  4  (/  +  15  =  0  to  (-  2,  3).  Ans.  -  %K 

49 

(d)  2  X  -  7  2/  +  8  =  0  to  (3,  -  5).  Ans 


+  V53 

12 

-  (e)  X  — 3i/ =  0  to(0,  4).  Ans.   -=. 

+  VIO 

2.  Do  the  origin  and  the  point  (3,  —  2)  lie  on  the  same  side  of  the 
line  X  —  y  +  1  =0?  Ans.    Yes. 

3.  Does  the  line  2x  +  3y  +  2  =  0  pass  between  the  origin  and  the 
point  (—  2,  3)  ?  Ans.   No. 

4.  Find  the  lengths  of  the  altitudes  of  the  triangle  formed  by  the 
lines  2x  +  32/  =  0,  x+32/+3  =  0,  and  x  +  y  +  1  =0. 

3  6  ,     /- 

Ans.    — z=5  — =1  and  V2. 
Vl3    VlO 

5.  Find  the  length  of  the  altitudes  of  the  triangles  whose  vertices  are 

(a)  (7,8),  (-8,4),  (-2,-10). 

(b)  (8,0),  (0,-8),  (-3,-3).  .^. 

(c)  (5,-4),  (-4,-5),  (0,8). 

6.  Find  the  equations  of  the  bisectors  of  the  angles  formed  by 

3x  —  42/4-1  =  0  and  ix-\-  Sy  —  1  =  0. 

Ans.    7  X  —  y  =  0  and  x+  7  y  —  2  =  0. 

7.  Find  the  locus  of  all  points  which  are  twice  as  far  from  the  line 
12x  +  5?/  —  l  =  Oas  from  the  ?/-axis.  Ans.   14x  —  5;/  +  l  =  0. 


THE  STRAIGHT  LINE  79 

8.  Find  the  locus  of  points  which  are  t  times  as  far  from  4  a;— 3  j/+ 1  =  0 
as  from  5x-12y  =  0.  Am.  (52  -  25k)  x  -  {39  -  60k)y  +  13  =  0. 

9.  Find  the  bisectors  of  the  angles  formed  by  the  lines  in  Problem  8. 

Ans.    77  X  -  09  2/  +  13  =  0  and  27  a-  +  21  y  +  13  =  0. 

10.  Find  the  distance  between  the  parallel  lines 

(a\/^  =  ^'^  +  ^'  "" 

^^>\y  =  2x-3. 

-*    ^°'    \y.=  -3x  +  i. 


Ans. 

V5 

Ans. 

3 

VTo 

Ans. 

1 

2\/T3 

Ans. 

6 

■^(c)  |2x-32/  +  4  =  0, 
^  '  \4x-6i/4-9=:0. 

/  2/  =  mx  +  3, 

\y  =  mx-3.  ""'■•   viT^ 

11.  Find  the  equations  of  the  bisectors  of  the  angles  of  the  following 

triangles,  and  prove  that  these  bisectors  meet  in  a  common  point : 

C?  (a)  X  +  2  ?/  -  5  =  0,         2  X  -  2/  -  5  =  0,  2  x  +  y  +  5  =  0. 

^(b)  3x+ 2/-1  =  0,  x-3y-3=:0,  x+3y+n=0. 

^(c)  3x  + 4?/-22  =  0,     4x-3?/  +  29  =  0,  y  -  5  -  0. 

(d)x  +  2  =  0,  2/- 3  =  0,  x  +  y  =  0. 

(e)  X  =  0,  2/  =  0,  x  +  y  +  3  =  0. 

12.  Find  the  bisectors  of  the  angles  formed  by  the  lines  4x  —  32/  —  1  =  0 
and  3x  —  4?/-f2  =  0,  and  show  that  they  are  perpendicular. 

Ans.    7 X  —  7y  +  1  =  0  and  x  +  y  —  3  =  0. 

13.  Find  the  equations  of  the  bisectors  of  the  angles  formed  by  the  lines 
5x-  12y  +  10  =  0  and  12x-  5y  +  1.5  =  0. 

14.  Find  the  locus  of  a  point  the  ratio  of  whose  distances  from  the  lines 
4x  — 3?/ '+ 4  =  OandSx  +  12)/- 8  =  0isl3to5.    Ans.  9x  +  9//-4  =  0. 

15.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by 
the#iiies  4x-3y  =  12,  5x-12?/-4  =  0,  and  12x  —  5?/ -  13  =  0. 
Show  that  they  meet  m  a  point. 

Ans.    7x-  9y-  16  =  0,  7x  +  7?/-  9  =  0,  112x-  64y-  221  =  0. 

16.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by 
the  lines  5x-12y  =  0,  5 x  +  12  2/  +  60  =  0,  and  12 x  -  5  y  -  60  =  0. 
Show  that  they  meet  in  a  point. 

Ans.    2  2/  +  -5  =  0,  1 7  X  +  7  2/  =  0,  17  X  -  17  //  -  60  =  0. 


80  NEW  ANALYTIC  GEOMETRY 

17.  The  sides  of  a  triangle  are  S x  +  4 y  —  12  =  0,  3x  —  4y  =  0,  and 
4x  +  3y  +  24  =  0.  Show  that  the  bisector  of  the  interior  angle  at  the 
vertex  formed  by  the  first  two  lines  and  the  bisectors  of  the  exterior  angles 
at  the  other  vertices  meet  in  a  point. 

18.  Find  the  equations  of  the  lines  parallel  to  3x  -f  4  ?/  —  10  =  0,  and 
at  a  distance  from  it  equal  numerically  to  3  units. 

-4ns.    oj  +  4?/ =  2.5  or  —  .5. 

:    35.  The  angle  which  a  line  makes  with  a  second  line.    The  angle 

between  two  directed  lines  has  Ix-rn  detiiied  (Art.  12)  as  the 

angle  between  their  positive  directions.    When  a  line  is  given 

by   means   of   its    equation,   no    positive 

direction  along  the  line  is  fixed.    In  order 

to  distinguish  between  the  two  pairs  of 

equal  angles  which  two  intersecting  lines 

make   with    each    other,    we'    define    the 

angle  which  a  line  makes  with  a  second  line 

to  be  the  positive  angle  (p.  2)  from  the 

second  line  to  the  first  line. 

Thus  the  angle  which  L^  makes  with  L,^  is  the  angle  6.  We 
speak  always  of  the  "  angle  which  one  line  makes  with  a  second 
line,"  and  the  use  of  the  phrase  "  the  angle  between  two  lines  " 
should  be  avoided  if  those  lines  are  not  directed  lines. 

Theorem.  If  vi^  and  m^  ai^e  the  slopes  of  tivo  lines,  then  the 
(ingle  0  which  the  first  line  makes  ivith  the  second  is  given  hy 

(VI)  tan  B  = 


1  +  m^m^ 


Proof  Let  ttj  and  a^  be  the  inclinations  of  L^  and  Z.,  respec- 
tively. Then,  since  the  exterior  angle  of  a  triangle  equals  the 
sum  of  the  two  opposite  interior  angles,  we  have 

(Fig.  1)  a^  =  $-\-  a,^,  or     e  =  n^-  a^, 

(Fig.  2)  a^  =  Tr-e  +  a^,     or     0  =  tt  +  (a^  -  a^.       ■ 


THE  STRAIGHT  LINE 


81 


And  since  (30,  p.  3),     tan  (tt  +  ^')  =  tana;, 
we  have,  in  either  case, 

tan  6  =  tan  (n^  —  a^ 

tan  aj  —  tan  a.^ 
1  +  tan  «!  tan  ag 


Y 

/l^ 

0 

/ 

X 

Fig.  1 


Fig.  2 


But  tan  a^  is  the  slope  of  Z^,  and  tan  or.,  is  the  slope  of  L- 
hence,  writing  tan  a^  =  m^,  tan  a,  =  m.^,  we  have  (VI). 

In  applying  (VI)  we  remember  that  m.^  =  slope  of  the  line 
from  which  B  is  measured  in  t\e,  positive  direction.  (The  Greek 
letter  B  used  here  is  named  "  theta.") 


EXAMPLES 

1.  Find  the  angles  of  the  triangle  formed  by  the  lines  whose  equa- 
tions are 

L:2x-3?/-6  =  0, 

M-.dx-y  -Q  =  0, 

JV:6x  +  42/-25  =  0. 

Solution.  To  see  which  angles  formed  by  the 
given  lines  are  the  angles  of  the  triangle,  we  plot 
the  lines,  obtaining  the  triangle  ABC. 

Let  us  find  the  angle  A.  In  the  figure,  A  is 
measured  from  the  line  L.  Hence  in  (VI),  m,  = 
slope  of  i  =  I,  m,  =  slope  of  M  =  6. 


.-.  tan  A  = 


_  16 
1  +  4~  15 


, and  A  =  tan-i||. 


82 


NEW   ANALYTIC   GEOMETRY 


Next  find  the  angle  at  B.    In  the  figure,  B  is  measured  from  N.    Hence 

in.,  =  slope  of  iV  =  —  L  nj,  =  slope  of  L  =  |.   Hence  m.,  = ,  and  B=  -■ 

Finally,  the  angle  at  C  is  measured  from  the  line  M.    Hence  in  (VI) 
m^  =  slope  of  M  =  6,  m^  =  slope  of  iV  =  —  ?. 
_  3  _  g       15 
•.  tan  C  =  —^ — ^  =  — ,  and  C  =  tan- 1  if. 


1-9        16 
We  may  vsrify  these  results.    For  if  B  ■ 


and 


- ,  then  A  =  1-C 

hence  (31,  p.  3  ;  and  26,  p.  3)  tan  ^  =  cotC  = r,,  which  is  true  for 

the  values  found. 

2.  Find  the  equation  of  the  line  through  (3,  5)  wliich  makes  an  angle 

of  -  with  the  line  x  —  y  +  Q 


0. 


Solution.    Let  m.^  be  the  slope  of  the  required  line.   Then  its  equation 
is  by  (II),  Art.  27, 
(1)  y-5  =  m^{x-3). 

The  slope  of  the  given  line  is  m.-,  =  1,  and  .since 
the  angle  which  (1)  makes  with  the  given  line  is 

~,  we  have  by  (VI),  since  0  =  ~  =  60°, 
3  o 

.-1 


3      1  +  )«^ 

??!,  —  1 


- 

7'/| 

/ 

/A 

A 

3 

\ 

/ 

-- 

_\ 

r3. 

'>) 

< 

I 

- 

. 

\ 

\ 

1 

\ 

0 

\ 

X 

\ 

Whence 


(2  +  V3). 


tan  — 
3 

1  +  J"l 

1  +  \^'3 

m.  = -: 

1  -  \3 
Substituting  in  (1),  we  obtain 

2/-5=-(2  + V3)(x-3), 

or  (2  +  V  3)  X  +  2/  -  (11  +  3  Vs)  =  0. 

In  plane  geometry  there  would  be  two  solutions  of  this  problem,  —  the 
line  just  obtained  and  the  dotted  line  of  the  figure.  Why  must  the  latter 
be  excluded  here  ? 

In  working  out  the  following  problems  the  student  should  first  draw 
the  figure  and  mark  by  an  arc  the  angle  desired,  remembering  that  this 
angle  is  measured  from  the  second  line  to  the  first  in  the  counterclock- 
wise direction. 


THE  STRAIGHT  LINE  83 

PROBLEMS 

1.  Find    the    angle    which    the    line    Sx  —  y  +  2  =0    makes    with 

2x  +  y  —  2  =  0;  also  the  angle  which  the  second  line  makes  with  the 

first,  and  show  that  these  angles  are  supplementary.  .         Sir    ir 

Ans.    — -1  —  . 
4      4 

2.  Find  the  angle  which  tlie  line 

(a)  2x—  5y  +  1  =  0  makes  with  the  line  x  —  2y  +  S  =  0. 

(b)  X  +  y  +  I  =  0  makes  with  the  line  x  —  y  +  1  =0. 

J  (c)  3x  —  4y  +  2  =  0  makes  with  the  line  x  +  Sy  —  7  =  0. 

(d)  6x  —  Sy  +  S  =  0  makes  with  the  line  x  =  6. 

(e)  X  —  7 ?/  +  1  =  0  makes  with  the  line  x  +  2?/  —  4  =  0. 

In  each  case  plot  the  lines  and  mark  the  angle  found  by  a  small  arc. 
Ans.    (a)  tan-i(-  Jj);    (b)  | ;      (c)  tan-i(V^)  ;      (d)  tan-i  (- ^)  ; 
(e)  tan- 1(3^5). 

^       3.  Find  the  angles  of  the  triangle  whose  sides  are  z  +  3?/  —  4  =  0, 
3x  —  2y  +  1  =  0,  andx  —  y  +  3  =  0. 

Ans.    tan-i(-  y),  tan-i{^),  tan-i(2). 

Hint.  Plot  the  triangle  to  see  which  angles  formed  by  the  given  lines  are 
the  angles  of  the  triangle. 

4.  Find  the  exterior  angles  of  the  triangle  formed  by  the  lines 
5x-j/  +  3  =  0,  2/  =  2,  x-42/  +  3  =  0. 

Ans.    tan- 1(5),  tan-i(-  |),  tan-i(-  V). 

6.  Find  one  exterior  angle  and  the  two  opposite  interior  angles  of 
the  triangle  formed  by  the  lines  2x  —  3y  —  6  =  0,  3x  +  4j/  —  12  =  0, 
x  —  Sy  +  6  =  0.    Verify  the  results  by  formula  37,  p.  3. 

i,'     6.  Find    the    angles   of    the    triangle    formed    by    3x+2j/  — 4  =  0, 
X  -  3  2/  +  6  =  0,  and  4  x  -  3  ?/  —  10  =  0. 

7.  Find  the  equation  of  the  line  passing  through  the  given  point  and 
making  the  given  angle  with  the  given  line, 

(a)  (2,  1),^,  2x-3?/  + 2  =  0.  Ans.    5x-y-9  =  0. 

4 

(b)  (1,  -  3),  -^,  X  +  22/  +  4  =  0.  Ans.    Sx  +  y  =  0. 

/  ^    ,  V  ,  ^  fn  +  tan  0  , 

(c)  (Xi,  Vi),  <p,y  =  mx  +  b.  Ans.   y-y^^  i_,^tan^  ^^  ~  ^^^' 


84  NEW  ANALYTIC   GEOMETRY 

36.  Systems  of  straight  lines.  An  equation  of  the  first  degree 
in  X  and.  y  which  contains  a  single  arbitrary  constant  will  repre- 
sent an  infinite  number  of  lines,  for  the  locus  of  the  equation 
will  be  a  straight  line  for  any  value  of  the  constant,  and  the 
locus  will  be  different  for  different  values  of  the  constant. 

The  lines  represented  by  an  equation  of  the  first  degree 
which  contains  an  arbitrary  constant  are  said  to  form  a  si/stem. 
The  constant  is  called  the  parameter  of  the  system. 

Thus  the  equation  y  =  2x  -\-  b,  where  h  is  an  arbitrary  con- 
stant, represents  the  system  of  lines  having  the  slope  2 ;  and 
the  equation  ?/  —  ,5  =  m(x  —  3),  where  m  is  an  arbitrary  con- 
stant, represents  the  system  of  lines  passing  through  (3,  5). 

The  methods  already  explained  suffice  for  solving  problems 
involving  straight  lines,  but  shorter  methods  result  in  some 
cases  by  using  systems  of  lines,  as  will  now  be  explained. 

Given  the  line 

(1)  3.T-f  2//-4  =  0. 
Now  every  line  of  the  system 

(2)  3.r  +  2//  =  7. 

is  parallel  to  (1),  for  the  slopes  of  (1)  and  (2)  are  equal. 
Again,  every  line  of  the  system 

(3)  .  2x-^y  =  k 

is  'perpendicidar  to  (1)  ;  for  the  slope  of  (3)  =  |,  the  negative 
reciprocal  of  the  slope  of  (1). 

Note  that  the  coefficients  of  x  and  y  in  (1)  and  (2)  are  the 
same,  while  the  coefficients  in  (1)  and  (3)  are  interchanged  and 
also  the  sign  of  one  of  them  is  changed. 

Next,  consid'er  the  line 

y-2=3(a;-f  2). 

It  passes  through  the  point  (—  2.  2).    Now  every  line  of  the 
system 

(4)  y-2=k{x-\-2) 


THE   STRAIGHT   LINE  85 

passes  through  this  point,  since  the  equation  is  satisfied  by  its 
coordinates  for  all  values  of  k. 

Again,  all  the  lines  in  the  system 
(5)  X  cos  k  +  //  sin  k  —  5  =  0 

are  at  a  distance  of  five  units  from  the  origin. 

The  value  of  the  parameter  k  will  depend  upon  the  condition 
imposed  upon  the  line  (2),  (3),  (4),  or  (5). 

Thus,  if  (2)  must  pass  through  (1,  —  3),  these  coordinates 
must  satisfy  (2),  and  hence 

3_6  =  A-.     .-.  k=-S. 
That  is,  the  equation  of  the  line  passing  through  (1,  —  3)  and 
parallel  to3.r-|-2y-4  =  0is3a;  +  2y/-f3  =  0. 

Again,  if  (4)  must  form  with  the  coordinate  axes  a  triangle 
of  unit  area,  we  set  one  half  the  product  of  its  intercepts  equal 
to  1.    Hence 

1/—  2  k  —  2\ 

or  k-^^'^^l=  0. 

^•.  k  =  -2,-^. 

Substituting  these  values  in  (4),  we  obtain 

2x  +  ?/  +  2  =  0,     a;  +  27/-2  =  0, 

both  lines  satisfying  the  above  conditions. 

Again,  if  (5)  must  pass  through  the  point  (10,  0),  then 

/ V3 

10  cos  k  =  5,  cos  k  =  ^,  sin  ^  =  ±  Vl  —  cos'^  k  =±  -p— ; 

and  substitution  gives  the  two  lines 

a:±  V3^-10  =  0. 
In  general,  we  may  say  this  :  In  finding  the  equation  of  a 
straight  line  defined  hy  two  conditions,  ire  Tnay  begin  by  writing 
down  the  equation  of  the  system  of  lines  which  satisfy  one  of 
these  conditions,  and  then  determine  the  value  of  the  parameter 
so  as  to  meet  the  second  condition. 


86  np:w  analytic  geometry 

PROBLEMS 

1.  Write  the  equations  of  the  systems  of  lines  defined  by  tlie  conditions 

(a)  Passing  through  (—  2,  3). 

(b)  Having  the  slope  —  |. 

(c)  Distance  from  the  origin  is  3. 

(d)  Having  the  intercept  on  the  ?/-axis  =—3. 

(e)  Passing  through  (6,  —  1). 

(  f  )  Having  the  intercept  on  the  x-axis  =  6. 

(g)  Having  the  slope  i. 

( h )  Having  the  intercept  on  the  2/-axis  =  5. 

( 1 )  Distance  from  the  origin  =  4. 

( j )  Having  one  intercept  double  the  other. 

(k)  Sum  of  the  intercepts  =  4. 

( 1  )  Length  intercepted  by  the  coordinate  axes  =  3. 

(ni)  Forming  a  triangle  of  area  6  with  the  coordinate  axes. 

2.  Determine  k  so  that 

(a)  the  line  2x  —  3?/  +  A;  =  0  passes  through  (—  2,  1).      Ans.  k  —  1. 

(b)  the  line  2fcc— 52/+3  =  0  has  the  slope  3.  Ans.  k  =  -V-. 

(c)  the  line  x  -{■  y  —  k  =  0  passes  through  (3,  4).  Ans.  k  =  7. 

(d)  the  line  Sx  —  4y  +  k  =  0  has  intercept  on  the  x-axis  =  2. 

Ans.    k  =  —  6. 

(e)  the  line  x  —  Sky  +4  =  0  has  intercept  on  the  ?/-axis  =—3. 

Ans.   A;  =—  |. 

(f )  the  line  ix  —  Sy  +  6k  =  0  is  distant  three  units  from  the  origin. 

A ns.    k  =  ±^. 

(g)  the  line  2x  +  7y  —  k  =  0  forms  a  triangle  of  area  3  with  tjie 
coordinate  axes.  Ans.    A:  =  ±  2  V2L 

3.  Find  the  equation  of  the  straight  line  which  passes  through  the  point 

(a)  (0,  0)  and  is  parallel  to  x  —  3  ?/  +  4  =  0.         Ans.    x  —  3  y  =  0. 

(b)  (3,  —  2)  and  is  parallel  to  x  +  ?/  +  2  =  0.       Ans.    x  +  y  —  1  =  0. 

(c)  (— 5,  6)  and  is  parallel  to  2x  +  4// —3  =  0.     Ans.    x  +  2y—7  =  0. 

(d)  (—  1,  2)  and  is  perpendicular  to3x  —  4?/  +  l  =  0. 

Ans.    4x  +  3;/  —  2  =  0. 

(e)  (—  7,  2)  and  is  perpendicular  tox— 3?/  +  4  =  0. 

Ans.    3x  +  2/  +  19  =  0. 

4.  The  equations  of  two  sides  of  a  parallelogram  ai'e  Sx  —  iy  +  6  =  0 
and  X  +  5y  —  10  =  0.  Find  the  equations  of  the  other  two  sides  if  one 
vertex  is  the  point  (4,  9).    Ans.    3x  —  47/  +  24  =  0  and  x  +  5 y  —  49  =  0. 


THE   STRAIGHT   LINE  87 

5.  Find  the  equation  of  the  straight  line  at  a  distance  of  three  units 
fronl  the  origin,  and  which  in  addition  satisfies  the  condition  given  : 

(a)  Parallel  to  the  line  2x  +  y  —  2  =  0.         Ans.    2x  +  y  ±SV5  =  0. 

(b)  Perpendicular  to  the  line  x  +  y  —  1  =  0.   Ans._x  —  y  ±S  a/2  =  0. 

(c)  Passing  through  the  point  (0,  8).      Ans.    ±  V55x  +  Sy  —  24:  —  0. 

(d)  Passing  through  the  point  (1,  5). 

(e)  Forming  a  triangle  with  the  coordinate  axes  whose  area  is  9. 

Stt 

(f )  Making  an  angle  of  —  with  the  line  x  +  2  y  +  4  =  0. 

*  Ans.    3x  +  2/ ±  3VTo  =  0. 

6.  Find    the    equation    of    the    straight    line    parallel   to   the    line 
3x4-4?/  — 7  =  0,  and  satisfying  the  additional  condition  given: 

(a)  Passing  through  the  point  (2,  —  6). 

(b)  Forming  a  triangle  of  area  2  with  the  coordinate  axes. 

(c)  Forming  a  triangle  of  perimeter  5  with  the  coordinate  axes. 

(d)  The  middle  point  of  the  intercepted  part  has  unit  abscissa. 

(e)  At  a  distance  of  three  units  from  the  origin. 

(f )  One  unit  nearer  to  the  origin. 

7.  Find  the  equation  of  the  line  passing  through  the  point  (3,  —  2) 
and  satisfying  the  additional  condition  : 

,  (a)   Parallel  to  the  line  x  —  7y  —  8  =  0. 

(b)  Perpendicular  to  the  line  3x  —  5^  =  7. 

(c)  Passing  through  the  point  (4,  1). 

(d)  Having  the  intercept  (—  7,  0). 

(e)  The  sum  of  its  intercepts  is  6. 

Ans.   x  —  3y  —  9  =  0,y  +  2x  —  4:  =  0. 

(f )  The  given  point  bisects  the  part  intercepted  by  the  coordinate  axes. 

(g)  Making  an  angle  of  45°  with  the  line  2x  —  Sy+  2  =  0. 

8.  Find  the  equation  of  the  line  parallel  to  the  line  3x  +  4?/  —  15  =  0, 
such  that  the  point  (2,  —4)  shall  lie  midway  between  the  two  lines. 

Ans.    3x  +  4y  +  S5  =  0. 

9.  Find  the  equation  of  the  straight  line  which  forms  a  triangle  of 
area  2  with  the  coordinate  axes,  and  whose  intercepts  differ  by  3. 

37.  System  of  lines  passing  through  the  intersection  of  two 
given  lines.    Given  the  two  lines 

(1)  L^:x  +  2!/-5  =  0, 

(2)  i„ :  3  X  -  v/  -  1  =  0. 


88  NEW  ANALYTIC   GEOMETRY 

Consider  the  system  of  lines  whose  equation  is 
(3)  .^  +  2  3/  -  5  +  A;  (3  x  -  2/  -  1)  =  0, 

where  /.;  is  an  arbitrary  number. 

It  is  easy  to  see  that  the  line  (3)  will  pass  through  the  in- 
tersection of  the  given  lines  L^  and  Z.,.  In  fact,  by  solving  (1) 
and  (2)  for  x  and  y,  we  find  a;  =  1,  y  =  2,  and  these  values 
satisfy  (3). 

Note  that  the  equation  (3)  is  formed  from  the  left-hand  mem- 
bers of  (1)  and  (2)  by  multiplying  one  of  them  by  the  parameter 
k  and  adding.  The  method  of  forming  (3)  shows  at  once  that 
the  line  it  represents  must  pass  through  the  intersection  of  the 
given  lines. 

Problems  requiring  the  equation  of  a  line  which  passes 
through  the  intersection  of  two  given  lines  are  often  much 
shortened  by  forming  the  equation  of  the  system  (3)  and  de- 
termining k  to  meet  the  given  condition.  The  advantage  of 
this  method  is  that  tve  do  not  need  to  know  the  coordinates  of 
the  point  of  intersection  of  L^  and  L^. 


EXAMPLES 

1.  Eind  the  equation  of  the  line  passing  through  P^  (2,  1)  and  the 
intersection  of  Xj :  3x  —  5?/  — 10  =  0  and  L^-.x -\-  y  -\-l —  0. 

Solution.  The  system  of  lines  passing  through  the  intersection  of  the 
given  lines  is  represented  by 

3x-5y-10  +  fc(x  +  ?/  +  1)  =  0. 
If  Pj  lies  on  this  line,  then 

6  -  5  -  10  +  i-  (2  +  1  +  1)  =  0  ; 
whence  fc  =  f . 

Substituting  this  value  of   k  and  sim- 
plifying, we  have  the  required  equation 
21 X- 11 2/ -31  =  0. 

2.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
i^:2x  +  ?/-l-l  =  0  and  L^:x  —  2y-\-\  =  (i  and  parallel  to  the  line  whose 
equation  is  L^  :  4  x  —  3  y  —  7  =  0. 


JHE   STRAIGHT   LINE 


89 


Solution.   The  equation  af  every  line  tlirouj^h  the  intersection  of  the 

first  two  given  lines  has  the  form 

2x  +  y  +  l  +  k(x-2y  +  l)  =  0, 

or       {2  +  k)x  +  {l-2k)y  +  {l  +  k)-^Q. 

2  +  k 

The  slope  of  this  line  is  — This 

^  1-2A- 

must  equal  the  slope  of  L., ;  that  is,  *. 

.-. =  :*,  or  A-  =  2. 

l-2k      ' 

Substituting  and  simplifying,  we  obtain 

4x  —  Sy  +  S  =  0.    Ans. 

Solve  the  followini,^  problems  without  iinding  the  point  of 
intersection  of  the  two  lines  given. 


PROBLEMS 

1.  Find  the  equation  of  the  line  passing  thruugh  the  intei-seotion  of 
2a;  —  37/4-2  =  0  and  S  x  —  4  ij  —  2  =  0^  and  which 

(a)  passes  througli  the  origin; 

(b)  is  parallel  to  5x  -  2  ?/  +  3  =  0 : 

(c)  is  perpendicular  to  3x  —  2  //  +  4  =  0. 

Ans.  (a)  5x  -  7  2/  =  0  ;    (b)  ox  -  2  y  —  50  =  0  :    (v)  2 x  +  Sy  -  -iS  =  0. 

2.  Find  the  equations  of  the  lines  which  pass  throtigh  the  vertices 
of  the  triangle  formed  by  the  lines  2x  —  Sy  +  l  =  i),  x  —  ij  =  0.  and 
3x4-4?/  —  2  =  0,  which  are 

(a)  parallel  to  the  opposite  sides; 

(b)  perpendicular  to  the  opposite  sides, 

A  n.s.   (a)  3  .r  4-  4  //  -  7  =  0,  14  a-  -  21 2/  4-  2  =  0,  17  X  -  17  (/  4-  5  =  0  ; 
(b)  4  X  -  3  ?/  -  1  =  0,  21 X  +  14  7/  -  10  =  0,  17  X  4-  17  //  -  9  =  0. 

3.  Find  the  equation  of  the  line  passing  through  the  intersection 
of  X  4- // —  2  =  0  and  x  —  y  +  (>  —  0  and  through  the  intersection  of 
2x  —  v  +  3  =  0  and  x  -  3  //  4-  2  =  0.  .1  ij.s.  l<)x  4-  3  //  4-  20  =  0. 

//////.  Th«  syntems  of  lint^s  ^Kissing  thr(mi;li  tht-  |)<iints  of  iutersec-tion  of  f  he 
two  pairs  of  HnoH  arc 

.r  +  .V  -  li  i  /•  \j:  -  ;i  +  (>)  =  0, 

and  2  X  -  ?/  +  3  +  A:'  ^  -  3  2/  +  2)  =  0. 


90  NEW  ANALYTIC   GEOMf:TRY 

These  lines  will  coincide  if  the  coefficients  are  proportional ;  that  is,  if 

2  +  k' ^  -\--Sk'~  3  +  2*'  ' 
Letting  r  be  the  common  value  of  these  ratios,  we  obtain 
1  +  A;  =  2  r  +  rk', 
\-k  =  -r-Zr¥, 
and  -2  +  Q>k='ir  +  2-rkf. 

From  these  equations  we  can  eliminate  the  terras  in  rk'  and  ;•,  and  thus  fiud 
the  value  of  k  which  gives  that  line  of  the  first  system  which  also  belongs  to 
the  second  system. 

4.  Find  the  equation  of  the  line  passing  through  the  intersection  <if 
2x  +  2/— 8  =  0  and  3 a;  +  2 r/  =  0  and 

(a)  parallel  to  the  ?/-axis.  Ans.   a;  —  16  =  0. 

(b)  parallel  to  the  x-axis.  Ans.    y  +  24  =  0. 

5.  The  equations  of  the  sides  of  a  parallelogram  are  a;  +  3?/4-2  =  0, 
X4.3y_8  =  0,  3x  —  2?/  =  0,  3x  —  2?/—  16  =  0.  Find  the  equations 
of  the  diagonals. 

6.  Find  the  equations  of  the  lines  through  the  point  of  intersection  of 
the  lines  x  +  Sy  —  10  =  0,  3x  —  2/  =  0,  which  are  at  unit  distance  from 
the  origin.  Ans.   x  —  1  =  0,  4x  —  3?/  +  5  =  0. 

7.  Find  the  equations  of  the  lines  through  the  point  of  intersection  of 
the  two  lines  7x4-  7?/  —  24  =  0,  x  —  ?/  =  0,  which  form  with  the  coordi- 
nate axes  a  triangle  of  perimeter  12. 

Ans.   4  X  +  3  y  -  12  =  0 ;  3x  +  4  ?/  -  12  =  0. 

REVIEW.    TRIANGLE  PROBLEMS 

1.  In  the  following  problems  the  coordinates  of  the  vertices  of  a  triangle 
are  given.  Find  (1)  the  equations  of  the  sides,  (2)  the  equations  of  the 
perpendicular  bisectors  of  the  sides,  (3)  the  equations  of  the  medians, 
(4)  the  equations  ©f  the  lines  drawn  from  the  vertices  perpendicular  to 
the  opposite  sides,  (5)  the  equations  oi-  the  lines  drawn  through  the  vertices 
parallel  to  the  opposite  sides,  (6)  the  lengths  of  the  three  medians,  (7)  the 
lengths  of  the  three  altitudes,  (8)  the  area,  (9)  the  three  angles,  (10)  the 
equation  of  the  circumscribed  circle. 

(a)  (8,  2),  (6,  6),  (-  1,  5).  (f )   (0,  -  4),  (6,  -  2),  (4,  -  5). 

(b)  (-  4,  5),  (-  3,  8),  (4,  1).  (g)  (-  3,  -  3),  (-  2,  0),  (5,  -  7). 

(c)  (4, 13),  (16,  5),  (-  1,  -  12).        (h)  (0,  2),  (8,  0),  (5,  5). 

(d)  (2,  4),  (8,  4),  (6,  0).  (i)  (3,  -  1),  (3,  -  5),  (0,  -  2). 

(e)  (4,  0),  (2,  4),  (-  5,  3).  ( j)  (-  1,  15),  (11,  7),  (-  6,  -  10). 


THE   STRAIGHT   LINE  91 

2.  In  the  following  problems  the  coordinates  of  the  vertices  of  a  triangle 
are  civen.  Find  (1)  the  equations  of  the  sides,  (2)  the  equations  of  the 
perpendicular  bisectors  of  the  sides,  (3)  the  equations  of  the  bisectors  of 
tlie  interior  angles,  (4)  the  equation  of  the  circumscribed  circle,  (5)  the 
equation  of  the  inscribed  circle. 

(a)  (8,  1),  (2,  4),  (-  2,  -  4). 

(b)  (6,  30),  (36,  -  10),  (-  24,  -  10). 

(c)  (3,  3),  (-  3,  6),  (-  7,  -  2). 

(d)  (0,  32),  (30,  _  8),  (-  30,  -  8). 

3.  In  the  following  problems  the  equations  of  the  sides  of  a  triangle 
are  given.  Find  (1)  the  angles,  (2)  the  equations  of  the  bisectors  of  the 
interior  angles,  (3)  the  equations  of  the  bisectors  of  the  exterior  angles, 
(4)  the  equation  of  the  inscribed  circle. 

(a)  4x-3?/-  4  =  0,  3x+42/-8  =  0,  5a;  -  12i/ -  60  =  0. 

(b)  5x  +  12  ?/  -  24  =  0,  12  X  +  5  ?/  +  7  =  0,  .5x  -  12  2/  -  48  =  0. 

(c)  5x-12  2/-42  =  0,  12x+  52/-2  =  0,  5x  +  12  ?/ -  66  =  0. 

(d)  12x+  6y+  50  =  0,  5x-12?/-81  =  0,  5x  +  12?/ -  33  r=  0. 
.   (e)  4x-3  2/  +  25  =  0,  5x-12?/+  1  =  0,  3x  +  4?/-5  =  0. 

(f)  5x  +  12  2/-123  =  0,  12x+  5?/  +  21  =  0,  5x  -  12  y  -  27  =  0, 

(g)  5x-12?/-3  =  0,  12x  +  5y  +  24  =  0,  5x  +  12  2/ -  75  =  0. 
(h)  12x+  5y  +  50  =  0,  5x+  12  2/-16  =  0,  5x-  12^-16  =  0. 


CHAPTER  V 

THE  CIRCLE 

38.  Equation  of  the  circle.  Every  circle  is  determined  when 
its  center  and  radius  are  known. 

Theorem.    The  equation  of  the  circle  whose  center  is  a  given 
Ijolnt  («,  /3)  and  whose  radius  equals  r  is 
(I)  (x-af  +  (y-pf  =  r\ 

Proof.    Assume  that  P(x,  y)  is  any  point  on  the  locus. 

If  the  center  (a,  /3)  be  denoted  by  C,  the  given  condition  is 

PC  =  r. 
By  (I),  p.  13,        PC  =  ^(x-af  +  {ij-pf. 
■■■  ^{x-af+{>j-iif  =  r. 
Squaring,  we  have  (I).  q.e.d. 

Corollary.  Tlis  equation  of  the  circle  whose  center  is  the  origin 
(0,  0)  and  whose  radius  is  r  is  ■ 

x^  -if-y^  z=  f-. 
If  (I)  is  expanded  and  ti-ansj^osed,  we  obtain 

(1)  a^  +  /  -  2  ^i:a;  -  2  )8y  +  a'  +  y8'  -  v'  =  0. 
The  form  of  this  equation  is  clearly 

2^"  4-  2/^  +  tei-ms  of  lower  degree  =  0. 

In  words :  Any  circle  is  defined  by  an  equation  of  the 
second  degree  in  the  variables  x  and  y,  in  which  the  terms  of  the 
second  degree  consist  of  the  stim  of  the  squares  of  x  and  y. 

Equation  (1)  is  of  the  form 

(2)  a-^  -f  y'  +  Dx  +  By  +  F  =  0, 
where 

(3)  D=-2a,E=-2p,  and  F  =  a^  +  1^  -  r\ 

92 


THE   CIRCLE  93 

Can  we  infer,  conversely,  that  the  locus  of  every  equation 
of  the  form  (2)  is  a  circle  ?  To  decide  this  question  transform 
(2)  into  the  form  of  (I)  as  follows:  Rewrite  (2)  by  collecting 
the  terms  in  x  and  the  terms  in  y  thus  : 

(4)  x^ -{- Dx  +  y""  +  Ey  =- F. 

Complete  the  square  of  the  terms  in  x  by  adding  (^-D)'^  to 
both  sides  of  (4),  and  do  the  same  for  the  terms  in  y  by  adding 
(i  E)'  to  both  members. 

Then  (4)  may  be  written 

(5)  (a-  +  \Df  +  {y  +  \Ef  =  \{D^+l^  -  4F). 

In  (5)  we  distinguish  three  cases  : 

If  Z)^  +  £'-  —  4Fis  positive,  (5)  is  in  the  form  (I),  and  hence 
the  locus  of  (2)  is  a  circle  whose  center  is  (—  ^D,  —  \E)  and 
whose  radius  is  r  =  ^  Vz*'^  +  is^  —  4  F. 

If  i)^  +  21^  —  4  F  =  0,  the  only  real  values  satisfying  (5)  are 
x——\D,y=—\E  (footnote,  p.  37).  The  locus,  therefore,  is 
the  single  point  (—  \D,  —  \E).  In  this  case  the  locus  of  (2) 
is  often  called  a  point  circle,  or  a  circle  whose  radius  is  zero. 

If  ly^  +  F^  —  4F  is  negative,  no  real  values  satisfy  (o),  and 
hence  (2)  has  no  locus. 

The  expression  If  -\-  E^  —  4,  F  \^  called  the  discriminant  of  (2), 
and  is  denoted  by  ©  (Greek  letter  "  Theta ").  The  result  is 
given  by  the 

Theorem.    TJte  locus  of  the  equation 
(II)  x'  +  y^-\.Dx  +  Ey  +  F=0, 

whose  discriminant  is  ®  =  Lf-\-  E^  —  A  F,  is  determined  as  foliates  : 

When  0  is  positive,  the  locus  is  the  circle  whose  center  is 
(—  1.  £>,  — ^  £•)  and  tvhose  radius  is  r  =  ^  Vy/-+7i'^— 4F=^  V®. 

When  0  is  zero,  the  locris  is  the  point  circle  (—  ^D,  —  ^-E). 

When  0  is  negative,  there  is  no  locus. 

Corollary.  When  E=0,  the  center  o/(II)  is  on  the  x-axLs,  and 
when  D=0,  the  center  is  on  the  y-axis. 


94 


NEW  ANALYTIC   GEOMETRY 


EXAMPLE 

Find  the  locus  of  the  equation  x^  +  y^  —  ix  +  8y  —  5  —  0. 
First  solution.   The  given  equation  is  of  the  form  (II),  where 
D  =  -4,   ^=8,   F  =  -5, 
and  hence 

e  =  16  +  64  +  20  =  100  >  0. 
The  locus  is  therefore  a  circle 
■whose  center  is  the  point  (2,  —  4) 
and  whose  radius  is  |  VlOO  =  5. 

Second  solution.    The  problem 

may  be  solved  without  applying  the 

theorem  if  we  follow  the  method  by 

which  the  theorem  was  established. 

Collecting  terms, 

(x2-4x)  +  (2/2  +  8y)  =  5. 
Completing  the  squares, 

(x2  _  4x  +  4)  +  (y2  +  8 y  +  16)  =  25. 
Or,  also,  (x  -  2)2  +  {y  +  4)2  =  25. 

Comparing  with  (I),      a:  =  2,  /3  =  —  4,  r  =  5. 

The  equation  Ax!^  +  Z>,r?/  +  Q/  +  Dx  -{-  Et/  -\-  F  =  0  is  called 
the  general  equation  of  the  second  degree  in  x  and  y  because  it  con- 
tains all  possible  terms  in  x  and  y  of  the  second  and  lower  de- 
grees. This  equation  can  be  reduced  to  the  form  (II)  when  and 
only  when  A  =  C  and  iJ  =  0.  Hence  the  locus  of  an  equation 
of  the  second  degree  is  a  circle  only  when  the  coefficients  of  x^ 
and  ?/  are  equal  and  the  xy-tenn  is  lacking. 

39.  Circles  determined  by  three  conditions.  The  equation  of  any 
circle  may  be  written  in  either  one  of  the  forms 

(x-ay^(y-/3y=r^, 
or  x^-\-y^  +  Dx  -i-  Ey  +  F=0. 

Each  equation  has  three  arbitrary  constants.  To  determine 
these  constants  three  equations  are  necessary.  Such  an  equa- 
tion means  that  the  circle  satisfies  some  geometrical  condition. 
Hence  a  circle  may  be  determined  to  satisfy  three  conditions. 


THE   CIRCLE 


95 


Rule  to  determine  the  equation  of  a  circle  satisfying  tltree 
conditions. 

First  step.    Let  the  required  equation  be 

(1)  (^^ay  +  {y-fif=r^, 
or 

(2)  x''  +  y^+Dx  +  Ey  +  F=0, 
as  may  be  more  convenient. 

Second  step.  Find  three  equations  betiveen  the  constants  a,  y3, 
and  r  \_or  D,  E,  and  F]  ivhich  express  that  the  circle  (1)  [o/'(2)] 
satisfies  the  three  given  conditions. 

Third  step.  Solve  the  equations  found  in  the  second  step  for 
a,  /3,  end  r  [_or  D,  E,  and  F^. 

Fourth  step.  Substitute  the  resxdts  of  the  third  step  in  (1)  [o/> 
(2)  ].    The  result  is  the  required  equation. 

In  some  problems,  however,  a  more  direct  method  results  by 
constructing  the  center  of  the  required  circle  from  the  given 
conditions  and  then  finding  the  equations  and  points  of  inter- 
section of  the  lines  of  the  figrure. 


EXAMPLES 

1.  Find  the  equation  of  the  circle  passing  through  the  three  points 
Pi(0,  1),  PjCO,  6),  and  P^(^,  0). 

First  solution.    First  step.    Let  the  re- 
quired equation  be 

(3)  x^  +  y^  +  Dx  +  Ey+  F=0. 
Second  step.    Since  Pj,  P^,  and  P^  lie 

on  (3),  their  coordinates  must  satisfy  (3). 
Hence  we  have 

(4)  1  +  E  +  F  =  0, 

(5)  36  +  6  £■  +  P  =  0, 
and 

(6)  9  +  3D+P=0. 

Third  step.    Solving  (4),  (5),  and  (6),  we  obtain 
E  =-7,     F=6,     D  =  -  5. 


96 


NEW   ANALYTIC   GEOMETRY 


Fourth  step.    Substituting  in  (3),  the  required  equation  is 

a;2  +  ?/-  —  5  X  —  7  y  +  6  =  0. 

The  center  is  (|,  |),  and  the  radius  is  f  V2  =  3.5. 

Second  solution.  A  second  method  which  follows  the  geometrical  con- 
struction for  the  circumscribed  circle  is  the  following.  Find  the  equations 
of  the  perpendicular  bisectors  of  P^P<,  and  P^P^.  The  point  of  intersec- 
tion is  the  center.   Then  tind  the  radius  by  the  length  formula. 

2.  Find  the  equation  of  the  circle  passing  through  the  points  P^  (0,—  3y 
and  P„(4,  0)  which  has  its  center  on  the  line  x  +  2y  =  0. 

First  solution.    First  step.   Let  the  required 
equation  be 

(7)  x2  +  if-  +  Dx  +  Ey  +  F=0. 

Second  step.    Since  Pj  and  P.^  lie  on  the  locus 
of  (7),  we  have 

(8)  9  -  3  7?  +  P  =  0, 
and 

(9)  16  +  4  Z;  +  F  =  0. 

, j,  and  since  it  lies  on  the  given  line, 

-f-(-D-. 

or 

(10)  D  +  2E  =  0. 

Third  step.    Solving  (8),  (9),  and  (10), 

U=-V,    F  =  l    P  =  -V- 
Fourth  step.    Substituting  in  (7),  we  obtain  the  required  equation, 

X2  +  2/2  _  y  X  +    7  y  _   2^4   =  0, 

or  5x2  + .5  2/2-14X  +  7?/-24  =  0. 

The  center  is  the  point  (|,  —  y'^),  and  the  radius  is  ^  V29. 

Second  solution.  A  second  solution  is  suggested  by  geometry,  as  follows  r 
Find  the  equation  of  the  perpendicular  bisector  of  PiPg.   The  point  of 
intersection  of  this  line  and  the  given  line  is  the  center  of  the  required 
circle.   The  radius  is  then  found  by  the  length  formula. 


Vi 

K 

/ 

■^ 

-vj 

S 

^ 

\ 

\ 

R, 

Ji 

0 

"^ 

(4,( 

)X 

\ 

4'^ 

& 

^ 

-4 

\ 

(0.-3) 

/ 

^ 

~^ 

P^ 

"^ 

1" 

THE   CIRCLE 


9T 


3.  Find  the  equation  of  the  circle  inscribed  in  the  triangle  whose 
sides  are 


(11) 


AB:3x-4y-l9  =  0, 

BC:ix  +  3y-n  =  0, 

CA:x+  7  =  0. 


Solution.  The  center  is  the  point  of  intersection  of  the  bisectors  of  the 
anu'les  of  the  triangle.  We  therefore  find  the  equations  of  the  bisectors 
of  the  angles  A  and  C. 

Reducing  equations  (11)  to        \C 
the  normal  form, 


(12)    AB: 


BC 


Sx-iy-19 


4x  +  3?/  —  17 


=  0: 


0; 


-1 

Then,  by  Example  2,  Art. 
34,  the  bisectors  ai-e 

3x-4?/-19_x+7 
2  .r  —  2/  +  4  =  0, 


(13)  ^D:- 


CE 


4x  +  3y  —  17      X  +  7 


5  -1 

or  3x  +  y  +  Q  =  0. 

The  point  of  intersection 
of  AI)  and  CE  is  (-  2,  0). 
This  is  therefore  the  center 
of  the  inscribed  circle.  The 
radius  is  the  perpendicular 
distance  from  any  of  the  lines  (11)  to  (—2,  0).  Taking  the  side  /I/', 
then,  from  (12), 

^^3(-2)-4(0)-19_      ^ 


Hence,  by  (I),  tlie  equation  of  the  required  circle  is 
(x  +  2)2  +  {!/-  0)2  =  25,     or  x'^  +  ?/  +  4x  -  21  =  0. 


Ans. 


98  NEW  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Find  the  equation  of  the  circle  whose  center  is 

(a)  (0,  1)  and  whose  radius  is  3.  Ans.  x^  +  y^  —  2y  —  8  =  0. 

(b)  (— 2,  0)  and  whose  radius  is  2.  Ans.  x^  +  y^  +4x  =  0. 

(c)  (—  3,  4)  and  whose  radius  is  5.  Ans.  x-  +  y^  +  6x  —  8y  =  0 

(d)  (a,  0)  and  whose  radius  is  a.  Ans.  x-  ■\-  y-  —  2ax  =  0. 

(e)  (0,  /3)  and  whose  radius  is  /3.  Ans.  x"  +  ?/-  —  2j8y  =  0. 

(f )  (0,  —  j8)  and  whose  radius  is  /3.  Ans.  x-  +  y^  +  2  ;8?/  =  0. 

2.  Draw  the  locus  of  the  following  equations  : 

(a)  x2  +  ?/2-6x-]6  =  0.  {i)x'  +  y'^-Qx  +  4?/ -5  =  0. 

(b)  3x2  +  3  ;/2  -  lOx  -  24  2/  =  q.  (g)  (x  +  1)2  +  {y  -  2)2  =  0. 

(c)  x2  +  ?/  =  8  X.  (h)  7  x2  +  7  7/2  _  4  X  —  2/  =  3. 

(d)  x2  +  ?/-  8x-Qy  -\-  25  =  0.  (i)  x2  + 1/2  + 2ax  + 26?/ 4-a^  +  ^'^  =  0. 

(e)  x2  +  2/2  -  2  X  +  2  2/  +  5  =  0.  (j )  x2  +  ?/2  +  10 x  +  100  =  0. 

3.  Show  that  the  following  loci  are  circles,  and  find  the  radius  and  the 
coordinates  of  the  center  in  each  case  : 

(a)  A  point  moves  so  that  the  s'.m  of  the  squares  of  its  distances  from 
(3,  0)  and  {-  3,  0)  always  equals  68.  Ans.   x'^  -\- y^  =  25. 

(b)  A  point  moves  so  that  its  distances  from  (8,  0)  and  (2,  0)  are  always 
in  a  constant  ratio  equal  to  2.  Ans.   x'^  +  2/2  =  16. 

(c)  A  point  moves  so  that  the  ratio  of  its  distances  from  (2, 1)  and  (  —  4. 2) 
is  always  equal  to  ^.  Ans.   3x^  -\-  3y"  —  24:X  —  4y  =  0. 

(d)  The  distance  of  a  moving  point  from  the  fixed  point  (—  1,  2)  is  twice 

its  distance  from  the  origin.  .  ,    ^  „  2  Vs 

Ans.    a  =  i,3  =  —2r  = 

(e)  The  distance  of  a  moving  point  from  the  fixed  point  (2,  —  4)  is  half 
its  distance  from  the  fixed  point  (0,  3). 

(f)  The  square  of  the  distance  of  a  moving  point  from  the  origin  is 
proportional  to  the  sum  of  its  distances  from  the  coordinate  axes. 

(g)  The  square  of  the  distance  of  a  moving  point  from  the  fixed  point 
(—  4,  3)  is  proportional  to  its  distance  from  the  line  3x  —  4?/— 5  =  0. 

(h)  The  sum  of  the  squares  of  the  distances  of  a  point  from  the  two  lines 
X  —  22/  =  0,  2x  +  2/  —  10  =  0,  is  unity. 

4.  Find  the  equation  of  a  circle  passing  through  any  three  of  the  fol- 
lowing points : 

(0,2)         (3,3)         (6,2)  (7,1)  (8,-2)  (7,-5) 

(6,-6)     (3,-7)     (0,-6)     (-1,-5)     (-2.-2)     (-1,1) 

Ans.   x2  +  ?/2  _  6  X  +  4  2/  —  12  =  0. 


THE   CIRCLE  99 

5.  Find  the  equation  of  tlie  circle  which 

(a)  has  the  center  (2,  3)  and  passes  through  (3,  —  2). 

Ans.   x^  +  y"^  —  ix  —  6y  —  13  =  0. 

(b)  has  the  line  joining  (3,  2)  and  (—  7,  4)  as  a  diameter. 

Ans.    ^2  + 2/2  + 4x  — 6?/  — 13  =  0. 

(c)  passes  through  the  points  (0,  0),  (8,  0),  (0,  —  6). 

A  ns.   a;2  +  2/2  —  8  X  +  6  2/  =  0. 

(d)  passes  through  (0,  1),  (5,  1),  (2,  -  3). 

Ans.    2 x2  +  2  2/2  —  10 x  +  2/  -  3  =  0. 

(e)  circumscribes  the  triangle  (4,  5),  (3,  —  2),  (1,  —  4). 

(f)  has  the  center  (—  1,  —  5)  and  is  tangent  to  the  x-axis. 

Ans.   x2  +  2/2  +  2 X  +  10 2/  +  1  =  0. 

(g)  has  the  center  (3,  —  5)  and  is  tangent  to  the  line  x  —  7y  +  2  =  0. 

Ans.   x2  +  2/2  —  6x  +  lOy  +  2  =  0. 

(h)  passes  through  the  points  (3,  5)  and  (—  3,  7)  and  has  its  center  on 

the  X-axis.  Ans.    x-  +  y"  +  ix  —  46  =  0. 

(i)  passes  through  ^the  points  (4,  2)  and  (—  6,  —  2)  and  has  its  center 

on  the  2/-axis.  Ans.   x^  +  y^  +  Sy  —  SO  =  0. 

(j)  passes  through  the  points  (5,  —  3)  and  (0,  6)  and  has  its  center  on 

the  line  2  X  — 3  2/  — 6  =  0.         Ans.    3x2  +  3 //- -  114x  -  64?/ +  27G  =  0. 

(k)  passes  through  the  points  (0,  2),  (—  1,  1)  and  has  its  center  in  the 

line  3 2/  +  2 X  =  0.  Ans.   x2  +  2/2  —  6x  +  4 2/  —  12  =  0. 

(1)  circumscribes  the  triangle  x  —  6  =  0,  x  +  22/  =  0,  x  —  22/  =  8. 

Ans.    2 x2  +  2  2/2  -  21 X  +  8  y  +  60  =  0. 
(m)  is  inscribed  in  the  triangle  (0,  6),  (8,  6),  (0,  0). 

Ai^s.   x2  +  2/2  —  4x  —  82/  +  16  =  0. 
(n)  passes  through  (1,  0)  and  (5,  0)  and  is  tangent  to  the  2/-axis. 

Ans.   x2  +  2/2  —  6 X  ±  2  Vs 2/  +  5  =  0. 
(o)  passes  through  the  points  (—  3,  —  1),  (1,  1)  and  is  tangent  to  the 
line  4x  +  3  2/ +  25  =  0.  -         ,  ^  /        ,iiA_Lr-   —   yi 

6.  Find  the  eijuations  of  the  inscribed  circles  of  the  following  triangles : 


(a)  x  +  2  2/  -  5  =  0, 

2x-2/  —  5  =  0, 

2X  +  2/+  5  =  0. 

(b)  3X  +  2/-1  =  0, 

X  -  3  2/  -  3  =  0, 

x  + 32/ +  11  =  0. 

(c)  3x  +  4?/-  22  =  0, 

4  X  -  3  2/  +  29  =  0, 

2/  -  5  =  0. 

(d)  X  +  2  =  0, 

2/  -  3  =  0, 

X  +  2/  =  0. 

(e)  X  =  0, 

y  =  o, 

X  +  2/  +  3  =  0. 

7.  What  is  the  equation  of  a  circle  whose  radius  is  10,  if  it  is  tangent 
to  the  line  4x  +  32/—  70  =  0at  the  point  whose  abscissa  is  10  ? 


100  NEW   ANALYTIC   GEOMETRY 

In  the  proofs  of  the  following  theoieins  the  choice  of  the  axes  of 
coordinates  is  left  to  the  student,  since  no  mention  is  made  of  either 
coordinates  or  equations  in  the  problem.  In  such  cases  always  choose 
the  axes  in  the  most  convenient  manner  possible. 

8.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
two  fixed  points  is  constant.    Prove  that  the  locus  is  a  cirde. 

9.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
two  fixed  perpendicular  line^  is  constant.   Prove  that  the  loous  is  a  circle. 

10.  A  point  moves  so  that  the  ratio  of  its  distances  from  two  fixed 
points  is  constant.    Determine  the  nature  of  the  locus. 

Ans.  A  circle  if  the  constant  ratio  is  not  equal  to  unity,  and  a  straight 
line  if  it  is. 

11.  A  point  moves  so  that  the  square  of  its  distance  from  a  fixed  point 
is  proportional  to  its  distance  from  a  fixed  line.  Show  that  the  locus 
is  a  circle. 


CHAPTER  VI 

TRANSCENDENTAL  CURVES  AND  EQUATIONS 

In  the  preceding  chapters  the  emphasis  has  been  laid  chiefly 
on  algebraic  equatioais  ;  that  is,  equations  involving  only  powers 
of  the  coordinates.  ^Ve  now  turn  our  attention  to  equations 
such  as  y  _  iQg  ^^     y  =  2^,     x  —  sin  ?/, 

which  are  called  transcendental  equations,  and  their  loci,  trany 
scendental  (•■urves. 

40.  Natural  logarithms.  The  common  logarithm  of  a  given 
number  N  is  the  exponent  x  of  the  base  10  in  the  equation 

(1)  10'  =  N ;  that  is,  x  =  log^^  .V. 

A.  second  system  of  logarithms,  known  as  the  natural  si/stew, 
is  of  fundamental  importance  in  mathematics.  The  base  of  this 
system  is  denoted  by  e,  and  is  called  the  natural  base.  Numer- 
ically to  three  decimal  places,  the  natural  base  is  always 

(2)  e  =  2.718. 

The  natural  logarithm  of  a  given  number  N  is  the  exponent  y 
in  the  equation 

(3)  ey  =  N;  that  is,  y  =  log,  N. 

To  find  the  equation  connecting  the  coinmon  and  natural 
logarithms  of  a  given  number,  we  may  take  the  logarithms  of 
both  members  of  (3)  to  the  base  10,  which  gives 

(4)  logj^  e«  =  logj„  N,  or  y  log^^  e  =  log^„  N.        (16,  p.  1) 

(5)  .".  logjij  .V  =  log^g  e  ■  logg  N  (using  the  value  of  y  in  (3)) 

101 


102 


NEW  ANALYTIC  GEOMETRY 


0.434;  also -  =  2.302. 


(A) 


The  equation  shows  that  the  common  logarithm  of  any  number 
equals  the  product  of  the  natural  logarithm  by  the  constant 
log  e.  This  constant  is  called  the  modulus  (=  il/)  of  the  com- 
mon system.    That  is  (Table,  Art.  2), 

<6)  M=\og^^ 

We  may  summarize  in  the  equations, 

Common  log  =  natural  log  times  0.434, 
Natural  log  =  common  log  times  2.302. 

These  equations  show  us  how  to  tind  the  natural  logarithm 
from  the  common  logarithm,  or  vice  versa. 

Exponential  and  logarithmic  curves.    The  locus  of  the  equation 

(7)  //  =  e^ 

is  called  an  exponential  curve.  From  the  preceding  we  may 
write  (7)  also  in  the  form 

(8)  a.  =  log,y  =  2.3021og^„y. 

The  locus  of  (7)  is  therefore  the  curve  whose  abscissas  are 
the  natural  logarithms  of  the  ordinates.  Let  us  now  discuss 
and  plot  (7).  (Figure,  p.  103.) 

Discussion.  Since  negative  num- 
bers and  zero  have  no  logarithms, 
y  is  necessarily  positive.  More- 
over, X  increases  as  y  increases. 
The  coordinates  of  a  few  points 
on  the  locus  are  set  dov?n  in  the 
table.  The  discussion  and  figure 
illustrate  the  fact  that 

logeO  =  —  00. 

For  clearly,  as  y  approaches  zero,  a;  becomes  negatively  larger  and  larger, 
without  limit.   Hence  the  x-axis  is  a  horizontal  asymptote. 

If  the  curve  is  carefully  drawn,  natural  logarithms  may  be  measured 
off.    Thus,  by  measurement  in  the  figure,  if 

2/ =  4,  x  =  1.38  =  loge4. 


.r 

0 

y 

X 

y 

1 

0 

1 

1 

e  =  2.7 

-1 

'-  =  .31 

e 

2 

i'-^  =7.4 

-2 

e^ 

etc. 

etc. 

etc. 

etc. 

TRANSCENDENTAL  CURVES  AND  EQUATIONS  103 


More  generally,  the  locus  of 

(9)  y  =  e^, 

where  A;  is  a  given  constant,  is  an  exponential  curve.  The  dis- 
cussion of  the  difference  of  this  locus  from  that  in  the  figure  is 
left  to  the  reader. 

The  locus  of  the  equation 

(10)  y  =  log,„  X, 
which  is  called  a  lo(j<irithniic  curve, 
differs  essentially  from  the  locus  of 
(7)  only  in  its  relation  to  the  axes. 
In  fact,  both  curves  are  exponential 
or  logarithmic  curves,  depending  upon 
the  point  of  view. 

The  locus  of  (10)  is  given  in  the 
figure  below.  Clearly,  since  log^^  0  = 
—  00,  the  y-axis  is  a  vertical  asymp- 
tote.   The  scales  chosen  are 

unit  length  on  A' A' '  equals  2  divisions, 
unit  length  on  YY'  equals  4  divisions. 


Fi 

L, 

(0 

I) 



— 

(1 

w; 

^ 

^ 

(3J,j 

) 

/ 

O 

/ 

(1, 

^) 

X 

r 

l(. 

h-i 

) 

Compound  interest  curve.   The  problem  of  compound  interest  intro- 
duces exponential  curves.   For,  if  r  =  rate  per  cent  of  interest,  and  n  = 
number  of  years,  then  the  amount  {=  A)  of  one  dollar  in  n  years,  if 
the  interest  is  compounded  annually,  is  given  by  the  formula 
4  =  (1  +  r)». 


104 


XEW  ANALYTIC  GEOMETRY 


For  example,  if  the  rate  is  5  per  cent,  the  formula  is 

(11)  .4  =  (1.05)". 

If  we  plot  years  as  abscissas  and  the  amount  as  ordinates,  the  corre- 
spondingcurve  will  beanexiwnentialcurve.  For,byArt.  2,  logiol.05  =  .021. 

Hence,  from  (A),  log^LOS  =  2.302  times  .021 

=  .048  (to  three  decimal  place.s). 
Hence,  by  (3),  e-o^^  =  1.05,  and  the  equation  (11)  becomes 

(12)  A  =  e-048n^ 
which  is  in  the  form  of  (9)  ;  that  is,  k  —  .048. 

For  convenience  in  plotting  exponential  curves  accurately 
the  following  table  is  inserted. 

Table  of  values  of  the  exponential  function  e^. 


X 

.0 

.1 

.2 

.3 

.4 

e-r  e-x 

e^      e-x 

ex      e-x 

ex     e-x 

ex     e-x 

0 

1 
2 
3 
4 
5 

1.00  1.00 
2.72  0.37 
7.39  0.14 
20.1  0.05 
54.0  0.02 
148  0.01 

1.11  0.90 
3.00  0.33 
8.17  0.12 
22.2  0.05 
00.3  0.02 
104  0.01 

1.22  0.82 
3.32  0.30 
9.03  0.11 
24.5  0.04 
06.7  0.01 
181  0.01 

1.35  0.74 
3.07  0.27 
9.97  0.10 
27.1  0.04 
73.7  0.01 
200  0.00 

1.49  0.07 
4.00  0.25 
11.0  0.09 
30.0  0.03 
81.5  0.01 
221  0.00 

X 

0 

.5 

.6 

.7 

.8 

.9 

ex      e-x 

ex      e-x 

ex     e-x 

ex      e-x 

ex     e-x 

1.05  0.01 

1.82  0.55 

2.01  0.50 

2.23  0.45 

2.46  0.41 

1 

4.48  0.22 

4.95  0.20 

5.47  0.18 

0.05  0.17 

6.09  0.15 

2 

12.2  0.08 

13.5  0.07 

14.9  0.07 

10.4  0.00 

18.2  0.06 

3 

33.1  0.03 

30.0  0.03 

40.4  0.02 

44.7  0.02 

49.4  0.02 

4 

90.0  0.01 

99.5  0.01 

110  0.01 

122  0.01 

134  0.01 

5 

245  0.00 

270  0.00 

299  0.00 

330  0.00 

305  0.00 

For  example,  to  find  the  value  of  e--^,  we  look  in  the  coUnnn  with  the 
caption  x  for  the  value  2  and  then  pass  to  the  right  under  the  caption  .3. 
The  value  sought  is  found  in  the  column  under  e^  to  be  9.97.  The  next 
value  to  the  right  of  this  under  e-^  is  e-2.3  —  o.lO. 


TRANSCENDENTAL  CURVES  AND  EQUATIONS  105 


PROBLEMS 
Draw  *  the  loci  of  each  of  the  following  : 


10.  y  =  e-^. 

PROBABILITY    CURVE 


7.  y  =  xe-  ^. 

8.  s  =  t^e-'. 

11.  y  =  21ogioX.         < 

12.  ?/  =  log,(l  +  x). 

13.  j/  =  21ogioix. 

14.  2/  =  logipVx. 

15.  2/  =  log«(l  +  e^). 

16.  s  =  log,o(l  +  2  0. 

17.  B  =  loge(l  +  i2). 

18.  x  =  logjo(l-  v/). 


41.  Sine  curves.  As  already  explained  (p.  2),  the  two  eoin- 
mon  methods  of  angular  measurement,  namely  circular  measure 
and  decree  measure,  employ  as  units  of  measurement  the  radian 
and  the  degree  respectively.  The  relation  between  these  unit*  is 
180 


(1) 


1  radian 


57.29  degrees, 

TT 

180' 
in  which  tt  =  3.14  (or  2_2  approximate!}),  as  usual 


1  degree  =  0.0174  radians     or 


H — i V 

-2-1.57  -1 


Degrees 

0° 


1     1.57  2 


^ 


— 7L  Radians  JL 

2  s 

Ecjuations  (1)  may  be  written  ' 

(2)  TT  radians  =  180  degrees. 

77"  77" 

Thus  :r  radians  =  90°,  --  radians  =  45°,  etc.   The  two  scales 
2  4 

laid  oft"  on  the  same  line  give  the  figure. 

*  If  the  shuiMi  only  of  the  curves  1-10  is  desired,  we  may  replace  e  by  the 
approximate  value  3. 


106 


NEW  ANALYTIC  GEOMETRY 


In  advanced  mathematics  it  is  assumed  that  circular  measure 
is  to  be  used.    Thus  the  numerical  values  of 

cos  — r- 

sm  2  X,     X  tan  —  >     

4  2  X 

for  X  =  1,  are  as  follows  : 

sin  2x  =  sin  2  radians  =  sin  114°.59  =  0.909, 
X  tan  — —  =  1  •  tan  |  -  radians )  =  tan  45°  =  1, 


irx  /TT 

■  cos  —       cos  I  —  radians 


cos  30° 


0.433. 


2.r  2  2 

Let  us  now  draw  the  locus  of  the  equation 
(3)  7/  =  sin;r, 

in  which,  as  just  remarked,  x  is  tlie  circular  measure  of  an  angle 


V    X 


Solution.  In  making  the  calculation  for  plotting,  it  is  convenient  to 
■choose  angles  at  intervals  of,  say,  30°,  and  tlien  find  x,  the  circular  measure 
of  this  angle,  in  radians,  and  y  from  the  Table  of  Ait.  4. 


Angle  in 

X 

Angle  in 

X 

degrees 

radians 

y 

degrees 

radians 

y 

0 

0 

0 

0 

0 

0 

30 

.52 

..50 

-    30 

J.-2 

-    .50 

60 

1.04 

.86 

-    60 

-1.C4 

-    .86 

90 

1.56 

1.00 

-    90 

-1.56 

-1.00 

120 

2.08 

.86 

-120 

-2.08 

-    .86 

150 

2.60 

.50 

-  150 

-  2.60 

-    .50 

180 

3.14 

0 

-180 

-  3.14 

0 

TRANSCENDENTAL  CURVES  AND  EQUATIONS  107 

Thus,  for  30°,  y  =  sin  30°  =  .50.  For  150°,  y  =  sin  150°=  sin  (180°-  30°)  = 
sin  30°  =  .50  (30,  p.  3). 

To  plot,  ciioose  a  convenient  unit  of  lengtli  on  XX'  to  represent  1  radian, 
and  use  the  same  unit  of  length  for  ordinates.  The  divisions  laid  off  on 
t!ie  X-axis  in  the  figure  are  1  radian,  2  radians,  etc.  Plotting  the  points 
(x,  y)  of  the  table,  the  curve  APOQB  is  the  result. 

The  course  of  the  curve  beyond  B  is  easily  determined  from  the 

relation 

sin(2  7r  +  x)  =  sinx. 

Hence  ?/ =  sinx  =  sin(2  7r  +  x); 

that  is,  the  curve  is  unchanged  if  x  +  2Tr  be  substituted  for  x.  This  means, 
however,  that  every  point  is  moved  a  distance  2  tt  to  the  right.  Hence 
the  arc  APO  may  be  moved  parallel  to  XX'  until  A  falls  on  B,  that 
is,  into  the  position  BRC,  and  it  will  also  be  a  part  of  the  curve  in  its 
new  position.  This  property  is  expressed  by  the  statement :  The  curve 
y  =  sinx  is  a  periodic  curve  with  a  period  equal  to  2  7r.  Also,  the  arc 
OQB  may  be  displaced  parallel  to  A"X' until  0  falls  upon  C.  In  this 
way  it  is  seen  that  the  entire  locus  consists  of  an  indefinite  number  of 
congruent  arcs,  alternately  above  and  below  XX'. 

General  discussion.  1.  The  curve  passes  through  the  origin,  since  (0,  0) 
satisfies  the  equation. 

2.  In  (3),  if     X  =  0,     y  =  sin  0  =  0  =  intercept  on  the  axis  of  y. 

Solving  (3)  for  x, 
(4)  X  =  arc  sin  y. 

In  (4),  if  2/  =  0,     X  =  arc  sin  0  =  nir,  n  being  any  integer. 

Hence  the  curve  cuts  the  axis  of  x  an  indefinite  number  of  times  botli 
on  the  right  and  left  of  O,  these  points  being  at  a  distance  of  tt  from 
one  another. 

3.  Since  sin(—  x)  =— sinx,  changing  signs  in  (3), 

—  y  =  —  sin  X, 
or  —  y  =  sin(—  x). 

Hence  the  locus  is  unchanged  if  (x,  y)  is  replaced  by  (—  x,  —  y),  and 
the  curve  is  symmetrical  with  respect  to  the  origin  (Theorem  II,  p.  43). 

4.  In  (3),  X  may  have  any  value,  since  any  number  is  the  circular 
measure  of  an  angle. 

In  (4),  y  may  have  values  from  —  1  to  +  1  inclusive,  since  the  sine  of 
an  angle  has  values  only  from  —  1  to  +  1  inclusive. 

5.  The  curve  extends  out  indefinitely  along  XX'  in  both  directions, 
but  is  contained  entirely  between  the  lines  ?/  =  +  1,  y  =—  1. 


108 


np:w  analytic  geometry 


The  locus  is  called  the  wave  curve,  from  its  shape,  or  the  sine  curve, 
from  its  equation  (3).   The  maximum  value  of  y  is  called  the  £implitude. 

Again,  let  us  construct  the  locus  of 

irx 
(5)  y  =  2sin  — . 

Solution.    We  now  choose  for  x  the  values  0,  J,  1,  1|,  etc.,  radians,  and 
arrange  the  vs'ork  of  calculation  as  in  the  table. 


X 

radians 

iTTX 

radians 

degrees 

.      -KX 

sin  — 
3 

y 

0 

0 

0 

0 

0 

i 

l-^ 

30 

.50 

1.00 

1 

iTT 

60 

.86 

1.72 

H 

4  7r 

90 

1.00 

2.00 

2 

JTT 

120 

.86 

1.72 

2^ 

i^ 

150 

.50 

1.00 

3 

■n 

180 

0 

0 

The  figure  represents  a  sine  curve  of  period  6  and  amplitude  2.  For 
the  curve  crosses  the  x-axis  ot  intervals  of  3,  and  the  maximum  value  of 
j/  equals  2.       v 


3X 


E(iuatiou  (5)  is  of  the  form 

1/  =  a  sin  Icji'. 
'1  he  iuuplitude  in  this  case  equals  a.    To  find  the  period,  set 

Jcx  =  'Zir.    Solvmg  for  x,  x  =  -—  =  period. 


TRANSCENDENTAL  CURVES  AND  EQUATIONS  109 

As  it  is  important  to  sketch  sine  curves  quickly,  the  follow- 
ing directions  are  useful : 

1.  Find  the  aurvplitude  and  the  jpercod. 

2.  Choose  the  same  scales  on  hath  axes. 

3.  L<rij  off  points  on  XX'  at  intervals  of  a  qua rter-period. 
The  hlgltest  and  loicest  points  are  at  the  odd  ([tiarter-perlods. 
The  intersections  with  A'A''  are  at  the  even  <iuarter-perlods. 

PROBLEMS 


Plot  the  loci  of  the  equations  : 
\,  *y  =  cosx  (see  figure). 


2.  y  —  sin  2x. 

Z.  y  —  cos  2  X 

4.  ?/  =  sin  \  X. 

5.  y  =  cos  ^x 

TT.r 
Q,  y  =  cos  — 


T.  y  =  sin 


8.  J/  =  3  cos  • 


9.  2/  =  3sin  — . 
5 


10.  y  =  2sin- 


11.  y  =  tanx. 
Y 


12.  y  =  secx. 


*  The  cosine  curve  differs  from  the  sine  curve  only  in  the  position  of  the 
y-axis.  The  highest  and  lowest  points  occur  at  half-periods  and  the  intersec- 
tions with  OX  at  odd  (quarter-periods. 


110 


NEW  ANALYTIC   GEOMETRY 


TTX 

17. 

y  =  cotx. 

21. 

?/  =  sec  i  X. 

13. 

y  =  tan 

4 

18. 

TTX 

22. 

2/  =  csc|x. 

14. 

2/  =  2  tan  x. 

w  =  cot 

''             4 

TTX 

IS. 

?/  =  2  tan 

23. 

TTX 

y  =  sec-—. 
4 

^                 3 

19. 

y  =  4  cot  — - . 

6 

TTX 

_ . 

TTX 

16.  2/  =  3  tan 


24.  2/ 


20.  y  =  cscx.  I    ^-^ 

25.  X  =  sin  ?/.    Also  written  y  =  arc  sin  x  or  sin-^x,   and  read  "the 
angle  whose  sine  is  x." 

26.  X  =  2  cos  y,  ovy  =  arc  cos  i  x. 

27.  X  =  tan  y,  or  y  =  arc  tanx  (see  figure). 

28.  X  =  2  sin  f  iry. 

29.  X  =  i  cos  1  Try. 

30.  y  =  arc  tan  ix. 

31.  2/  =  2  arc  cos^x. 

The  locus  of  the  equation 


Yj 

^ 

^ 

/ 

___.^^ 

0             A 

__^ 

^ 

<6) 


?/  =  J  siu 


7r.r       TT 
T  +  6 


is  also  a  sine  curve.   For,  by  taking  the  coefficient  of  x,  namely  — j 
•outside  the  parenthesis,  (6)  becomes 


?/  =  i  sni  tt(  .^  +  7 


<8) 


Kow  set  a-  +  1  =  a;'.    Substituting  in  (7),  the  latter  becomes 
y  =  2  sni— — • 

But  this  is  equation  (5)  except  that  x'  takes  the  place  of  x. 
Hence  to  draw  the  locus  of  (6),  proceed  thus  :  Mark  the  point 
ic  =  —  ^  (or  x'  =  0)  on  the  .r-axis.  Using  this  point  as  the  new 
origin,  plot  the  locus  of  equation  (8). 

The  figure  obtained  is  obviously  precisely  that  on  page  108, 
if  the  y-axis  is  moved  to  the  right  a  distance  equal  to  J. 

Observe  that  the  period  of  (6)  is  determined,  as  before,  by 


the  coefficient  of  x.    The  added  term 
intercepts  on  the  cc-axis. 


simply  affects  the 


TRANSCENDENTAL  CURVES  AND  EQUATIONS     111 

PROBLEMS 

Plot  the  curves : 

1.  y  =  sm{z  +  i).  1.  y  =  ism(ix  +  l). 


2.  2/  =  2  cos  (2  X  —  1) 

('2  7rx 


y  =  2  sin  (  ^^  + -\ 


8.  y  =  cos  (x  +  ^j. 
3      ■    4/  9.  s  =  asin{kx  +  tt). 


4.  2/  =  2sin(2  7rx  +  ^V  10.  x  =  2  cos  (^  - -V 

/ttx      2  7r\  ,,  .    /2  7JT/         \ 

5.  w  =  3  cos 1 .  11.  X  =  sin  (  — ~  —  •n-)  • 

\2         3/  \   3  / 

6.  y  =  tan  /  —  +  - j  .  12.  s  =  a  cos  /  —  +  /3j  • 

42.  Addition  of  ordinates.  When  the  equation  of  a  curve  has 
the  form 

y  =  the  algebraic  sum  of  two  expressions, 

as,  for  example,  y  =  sin  x  +  cos  x,  y  =  \x  -\-  sin^  x,  s  =  e*  -\-  e~ ', 
etc.,  the  principle  known  as  addition  of  ordinates  may  with  ad- 
vantage be  employed.    For  example,  to  construct  the  locus  of 

(1)  y  =  2sm  —  +  -x, 
we  employ  the  auxiliary  curves 

(2)  y,=  2sin^, 

Plot  these  curves  one  below  the  other,  keeping  the  y-axes  in  a 
straight  line.  The  same  scales  must  be  used  in  both  figures. 
The  locus  of  (2)  is  the  sine  curve  of  Fig.  1,  p.  112.  The  locus 
of  (3)  is  the  straight  line  in  Fig.  2. 

The  ordinates  of  Fig.  1  are  now  added  to  the  corresponding 
ones  in  Fig.  2,  attention  being  given  to  the  algebraic  signs. 
The  derived  curve  A^B^OB^A,^  has  the  equation 

iV  2/  =  yi  +  y,=  2sni— +  -a; 


112 


XEW  ANALYTIC   GEOMETRY 


as  required.  The  locus  winds  back  and  forth  across  the  line 
2/  =  I  .r,  crossing  the  line  at  a;  =  0,  ±4,  ±8,  ±  12,  etc. ;  that 
is,  directly  under  the  points  where  the  sine  curve  in  Fig.  1  crosses 
the  ^--axis. 


Fig.  1 


Fig.  2 


PROBLEMS 

Draw  the  following  curves  and  calculate  y  accurately  for  the 
value  of  X : 

1.  2/  =:  cosx  +  Ix.         X  =  1. 


given 


2.  2/  =  sin2x  + 


10 


3.  y  =  sinx  +  cosx.       x  =—  ^. 

4.  y  =  -x—3sin x  =  2. 

4  3 


X-"  .  TTX 

5.  y  = 4  cos  — 

16  4 


x=-2. 


6..-        ^        . 

X 

7.  2/  =  e^  —  sin2x. 

X 

e>  -  e- « 
ft    11  — 

2 

9.  y  =  e*  —  cos4x.         x  =  ^tt. 
10.  y  =  sin  x  +  sin  2  x.    x  =  0.8. 


TRANSCENDENTAL  CURVES  AND  EQUATIONS  113 


.       TTX                    TTX 
11.   U  =  Sin h  t-os X 

4  3 


12.  J/  =  sin  ax  4-  cos  ax.   x  =  -^^  a. 

13.  ?/  =  2sinx  +  5cosx.    x  =  0.5. 

14.  y  =  2  sin  2  x  +  3  cos  |  x.    x  =  2. 

15.  ?/  =  sin  ax  +  sin  hx. 

16.  y  =  VO  —  X-  +  sin  27rx.    x  =  2^. 

17.  2/  =  e-^  +  4x-.   X  =—  2.4. 

•      2  TTX 

18.  y  =  logipX  +  sm  ^- .  x  =  2 

/-  1  /tTX       •  TT 

19.  w  =  2  Vx  4-  -  cos    —  +  - 

2        \  2        3 


20.  y  =  -{e"  +  e    a). 


2\a. 


The  locus  in  Problem  20  is  called 
the  catenary  (see  figure).  The  shape 
of  the  curve  is  that  as.sumed  by 
a  heavy  flexible-  cord  freely  sus- 
pended from  its  extremities. 


The  student  may  have  observed  from  the  preceding  exam- 
ples the  truth  of  the  following 

Theorem.  The  curve  obtained  by  adding  corresj^onding  ordi- 
nates  of  sine  curves  with  the  same  period  is  also  a  sine  curve 
with  eqtial  period. 

For  example,  consider  the  equation 


(5) 


/2  7rt        \  /2'irt         \ 

y  =  asml-^  -faj+/>sin(—  +/31, 


2'7rt 


in  which  a,  /8,  and  P  are  constants.    The  period  of  both  sine 
curves  equals  P.    Expand  the  right-hand  member  by  the  rule 

(33,  p.  3)   for  sin  {x  +  y)   and  collect  the  terms  in  sin 

2  irt 
and  cos Then  equation  (5)  assumes  the  form 

.    2'Trt  2  irt  Cy 

(6)  //  =  A  sm  — —  +  B  cos  -— -  ? 

where  A  and  B  are  constants,  independent  of  t. 

Let  us  now  introduce  the  angle  y  of  the  right  triangle  whose  legs 

are  A  and  B.  Let  the  hypotenuse  V.4^  +  B^  =  C.  Then  B  —  C  sin  y, 


114 


NEW  ANALYTIC  GEOMETRY 


A  =C  cos  y.    Substituting  these  values  in  (6)  gives 

y-      -^  COS  y  +  cos  — 


(()    ?/  =  C[sin  —^  cosy  +  cos— ^  smy  1  =  C  sinl  — --  +  y  )• 


This  is  a  sine  curve  with  period  P  and  amplitude  C  =  '\/ A^-\-E^. 

Q.E.D. 

The  curve  resulting  from  the  addition  of  ordinates  of  sine 
curves  with  unequal  periods  is,  however,  not  a  sine  curve. 

43.  Boundary  curves.    In  plotting  the  locus  of  an  equation 
of  the  form 
(1)  y  =  jjroduct  of  two  factors 

one  of  which  is  a  sine  or  cosine,  as,  for  example, 

TTt 

y  =  e'  sm  x,     or     s  =  t'  cos  —  ? 

much  aid  is  obtained  by  the  following  considerations : 
For  example,  consider  the  locus  of 


(2) 


y  =  e^   sm— -• 


We  now  make  the  following  observations  : 

1.  Since  the  numerical  value  of  the  sine  never  exceeds  unity, 
the  values  of  ij  in  (2)  will  not  exceed  in  numerical  value  the 
value  of  the  first  factor  e~^^.  Moreover,  the  extreme  values  of 
sin  -^  TTX  are  + 1  J^i^d  —  1  respectively.    Hence  y  has  the  extreme 

vaiues  e   *    and  —  e   *  . 

Consequently,  if  the  curves 

(3)  y  =  e~^^    and     y=-e-*" 

are  drawn,  the  locus  of  (2)  will  lie  entirely  between  these  curves. 

They  are  accordingly  called  boundary  curves. 

Draw  these  curves.  The  second  is  obviously 
symmetrical  to  the  first  with  respect  to  the 
a:-axis.  To  plot,  find  three  points  on  the  first 
curve,  as  in  the  table.    (Use  the  Table,  p.  104.) 


X 

y 

0 

2 
4 

1 

e-  ^  =  .61 
e-i  =  .37 

TRANSCENDENTAL  CURVES  AND  EQUATIONS  115 

2.  When  sin  ^ttx  =0,  then  in  (2)  y  =  0,  since  the  first  fac- 
tor is  always  Jinlte.  Hence  the  locus  of  (2)  vieets  the  x-axls  In 
the  same  points  as  the  auxiliary  sine  curve 

(4)  y  =  sin  ^  ttx. 

3.  The  required  curve  touches  *  the  boundary  curves  when  the 
second  factor,  sin  i  ttx,  is  -|-  1  or  —  1 ;  that  is,  when  the  ordinates 
of  the  auxiliary  curve  (4)  have  a  maximum  or  minimum  value. 

Hence  draw  the  sine  curve  (4).  The  period  is  4  and  the 
amplitude  is  1.    This  curve  is  the  dotted  line  of  the  figure. 


The  discussion  shows  these  facts : 

The  locus  of  (2)  crosses  XX'  at  x  =  0,  ±  2,  ±  4,  +  6,  etc.,  and 
touches  the  boundary  curves  (3)  at  x  =  ±  1,  ±3,  ±,5,  etc. 

We  may  then  readily  sketch  the  curve,  as  in  the  figure ;  that 
is,  the  winding  curve  between  the  boundary  curves  (3). 

4.  For  a  check  remember  that  the  ordinate  of  (2)  is  the 
product  of  the  ordinates  of  the  boundary  curve  y  =  e~^^  and 
the  sine  curve  (4).  In  the  figure,  for  example,  the  required 
curve  lies  above  A'A''  between  x  =  0  and  x  —  2,  for  the  ordinates 
of  7j  =  e~^^  and  of  the  sine  curve  are  now  all  positive.  But 
between  x  =  2  and  .x  =  4  the  required  curve  lies  below  A'A'',  for  the 
ordinates  of  y  =  e~^'^  and  the  sine  curve  now  have  unlike  signs. 

*The  discussion  shows  merely  that  the  curve  (2)  reaches  the  boundary 
<5urves.    Tangency  is  shown  by  calculus. 


116  NEW  ANALYTIC   (JEO.METRY 


PROBLEMS 

Draw  the  following  loci  and   calculatt-   y  accurately  for  the  given 
values  of  x  : 

^   ■  n      ,  ,,  sinx/     ]    .      \ 

1.  y  =  -  smx.  x  =  2;   lir.       11.  i/ = (=-.siiix|.    x=0.1. 

4  "  X    \     X         J 

r2  ,  „  sin  2x  „  ,     , 

2.  i/  =  — cos2x.  j;  =  l;7r.  12.?/=:— j-  =  0.1;l. 

cos  X 

^.ir.'^^  .  _  Q  .     1  13.    //  =  —    .       X  =    1  ;    TT. 


3.  7/  =  -  sin X  =  S  ;  I. 

■'3         3 


A  X-  TTX  o       oi 

4.  «  =  —  cos X  =  3  ;  2 ,', . 

10         5 


X 


U.y  =  ^^.    .  =  0.2;  r 
x^ 


.      TTX 

5.  y  =  e-^sinx.  x  =^Tr:  ^ir.  sin -^ 


15.  y  = X  =0.1  ;  2. 


B.  y  =  e-^cos  2x.  x  =  J  tt  ;  2. 

7.  v/  =  e    -    Sin x  =  —  2  ;   .•.       , .  .1  „  , 

4  16.  //  =  sm  -  X  cos  2  x .    x  =  ^  tt 

8.  w  =  e^i'^cos!!^  3._Q._i  /        ]\        1 

3-  ^-^'       ^--      17.  ,/  =  (x  +  -)sin-a;. 

.—It  /tX  7r\  o  1  ■$  -, 

cos  I  - 


cos-x COS -a;. 

4        2         4       2 


-  u-x  /2  TTX  \  ,„  _/     .  ,  _!/     .      7rf 

10.  y  =  ae         cos  ( !-«)•  19-   '/ =  c      suiirt  +  e   ^   sin  — 

'    P  /  2 

20.  Draw  the  two  loci  obtained  (1)  l)y  adding  and  (2)  by  multiplying 

the  ordiiiates  in  the  following  pairs  of  curves  : 

C                 x2 
f  _?^  I  ?/  =  2  H , 

y  =  cos 


(c     ^  i>  -^     '  (e)  < 

U  =  sinx.  [2/  =  sin7r./'.  |  ^.^^os  — 

I  '3 


r 

16 

-  X2 

y 

— 

— 

8 

(t) 

' 

TTX 

•II 

— 

COS 

I 

2 

2/  =  3  +  —. 
(b)  ^  ?/  =  e^  (d)  j  J^ 

1^.(7  =  cos7r.f.  I  y  =  sin-  - 

l  2 

44.  Transcendental  equations.  Graphical  solution.  The  solution 
of  certain  equations  of  frequent  occurrence  may  be  simplified 
by  using  graphical  methods. 


TRANSCENDENTAL  CURVES  AND  EQUATIONS  117 


Consider  the  equation 

(1)  cot  X  =  X,  or  cot  X  —  X  :=  0. 
To  find  values  of  x  {in  radians) 

for  which  this  equation  holds. 

To    aid    in    determining    the 
roots,  let  us  draw  the  curves 

(2)  y  =  cot  X   and    y  =  x. 
Now  the  abscissa  of  each  point 

of  intersection  is  a  root  of  equa- 
tion (1),  for,  obviously,  at  each 
point  of  intersection  of  the  curves 
(2)  we  must  have  cot  x  =  x;  that 
is,  equation  (1)  is  satisfied. 

In  plotting  it  is  well  to  lay  off  carefully  both  scales  (degrees 
and  radians)  on  OX. 


y  =cot  X 

Degrees 

X 

radians 

y 

0 

0 

CO 

10 

.174 

.5.67 

20 

.342 

2.75 

30 

.524 

1.73 

40 

.698 

1.19 

45 

.785 

1.000 

50 

.873 

.839 

60 

1.047 

.577 

70 

1.222 

.364 

80 

1.396 

.176 

90 

1.571 

.0 

1§0  190  200°Jf 


Number  of    solutions.    The   curve   y  =  cot  x  consists   of   an 
infinite  number  of  branches  congruent  to  AQJJ  of  the  figure. 


118  NEW  ANALYTIC   GEOMETRY 

The  line  y  =  x  will  obviously  cross  each  branch.    Hence  the 
equation  (1)  has  an  infinite  number  of  solutions. 

Smallest  solution.  From  the  figure  this  solution  lies  between 
45°  and  50°,  or,  in  radians,  between  x  =  .785  and  x  =  .873. 
Hence  the  first  significant  figure  of  the  smallest  root  is  0.8. 
Inter];)olation  is  necessary  to  determine  subsequent  figures. 

For  this  purpose  arrange  the  work  thus,  using  the  preceding 

table.  X  (radians)  cotx  cotx  —  x 

.873  .839  -.034 

.785  1.000  +.215 

difference     +  .088  -  .249 

We  wish  to  know  what  change  in  x  above  .785  will  produce 
a  decrease  in  cot  x  —  x  equal  to  .215 ;  that  is,  make  cot  x  —  x 
equal  to  zero.    Call  this  change  z.    Then,  by  proportion, 

:o88  =  3:249'    ■■'  =  -^'^ 

Hence  x  =  .785  +  .076  =  .86  (to  two  decimal  places). 

PROBLEMS 

Determine  graphically  the  number  of  solutions  in  each  of  tlie  following, 
and  find  the  smallest  root  (different  from  zero). 

Ans.    One  solution;  x  =  0.74. 
Ans.   Three  solutions;  x  =  0.95 
Ans.   Infinite  number. 
Ans.   Three. 
Ans.   Two. 
Ans.   Two. 

13.  3sinx  =  2cos4x.  19.  e^  =  tanx.  , 

14.  2  sin  -  =  cos  2  X.  ^0.  sin  x  =  log^,,  x. 
2 

21.  cosx  =  logjgX. 

15.  sinSx  =  cos2x. 

16.  e-  =  x.  22.  tanx  =  log,ox. 

17.  e^  =  sinx.  23.  e-' =  log«x. 

18.  e-*  =  cosx.  24.  e-^'  =  x^. 


1. 

cos  X  =  X. 

2. 

sin2x  =  X. 

3. 

tan  X  =  X. 

4. 

sinx  =  |x. 

5. 

sinx  =  x^. 

6. 

cosx  =  x^. 

7. 

tanx  =  x^. 

8. 

cotx  =  x^. 

9. 

X 

cosx=  -. 
3 

10. 

tan  X  =  1  —  X. 

11. 

cosx  =  1  —  X. 

12. 

3sinx  =  cosx 

CHAPTER  VII 


POLAR   COORDINATES 


0\ 


.  45.  Polar  coordinates.    In  this  chapter  we  shall  consider  a 

second  method  of  determining  points  of  the  plane  by  pairs  of 

real  numbers.    We    suppose  given  a       p 

fixed  point   0,  called    the   pole,    and 

a  fixed  line   OA,  passing  through   0, 

called  the  polar  axis.    Then  any  point 

P  determines  a  length  OP  =  p  (Greek 

letter  "  rho  ")  and  an  angle  AOP  =  0.  \ 

The  numbers  p  and  6  are  called  the  \ 

polar  coordinates  of  1\    p  is  called  the 

radius  vector  and  0  the  vectorial  angle.    The  vectorial  angle  6  is 

jjositive  or  negative  as  in  trigonometry.    The  radius  vector  is 

positive  if  P  lies  on  the 

terminal  line  of  0,  and  ,       \  ,       ,,, 

negative  if  P  lies  on  that 
line  produced  through 
the  pole  O. 

Thus  in  the  figure  the 
radius  vector  of  P  is 
positive,  and  that  of  P' 
is  negative. 

It  is  evident  that 
every  pair  of  real  nnm- 
hers  (p,  6)  determines  a 
single  point,  which  may 
be  plotted  by  the 

"Rule  for  plotting  ajjoint  whose  jjolar  coordinates  (p,  6^  are  given. 

119 


120  NEW  ANALYTIC   GEOMETRY 

First  step.  Construct  the  terminal  line  of  the  vectorial  angle  0, 
as  in  trigonortietry. 

Second  step.  If  the  7'adius  vector  is  positive,  lay  off  a  length 
OP  =  p  on  the  terminal  line  of  6 ;  if  negative, prodtice  the  termi- 
nal line  through  the  pole  and  lay  off  OP  equal  to  the  numerical 
value  of  p.    Then  P  is  the  required  p)oint. 

In  the  tigure  on  page  119  are  plotted  the  points  whose  polar 
coordinates  are  (6,  60°),  (s,  ^) ,  (-  3,  225°),  (6,  180°),  and 

Every  point  determines  an  inf- 
nitenumber  of  pairs  of  numbers  (p,  &). 

Thus,  if  OB  =  p,  the  coordinates 
of  B  may  be  written  in  any  one  of 
the  forms  (p,  6),  (-  p,  180°  +  0), 
(p,  360°  +  e),  (~p,6-  180°),  etc. 

Unless  the  contrary  is  stated,  we  shall  always  suppose  that 
$  is 2>ositive,  or  zero,  and  less  than  360°;  that  is,  0^6<360°. 

.    PROBLEMS 


<5,  ,r). 


Plot  the  points  (4,  45°),  (6,  120°),  i-  2,— V  U,  ^V  (-  4,  -  240°), 


2.  Plot  the  points  U,  ^  -\  (-2,  ±  ^Y  {3,7r),  (-  4,7r),  (6,  0),  (-  6, 


0). 


3.  Show  that  the  points  (p,  6)  and  (p,  —  d)  are  symmetrical  with  respect 
to  the  polar  axis. 

4.  Show  that  the  points  (p,  6),  {—  p,  ff)  are  symmetrical  with  respect 
to  the  pole. 

5.  Show  that  the  points  (—  p,  180°  —  0)  and  (p,  0)  are  symmetrical  with 
respect  to  the  polar  axis. 

46.  Locus  of  an  equation.  If  we  are  given  an  equation  in  the 
variables  p  and  0,  then  the  locus  of  the  equation  is  a  curve  such  that 

1.  Every  point  whose  coordinates  (p,  6)  satisfy  the  equation 
lies  on  the  curve. 


POLAR   COORDINATES 


121 


2.  The  coordinates  of  every  point  on  tlie  curve  satisfy  the 
eqx:ation. 

The  curve  may  be  plotted  by  solving  the  equation  for  p  and 
finding  the  values  of  p  for  particular  values  of  $  until  the 
coordinates  of  enough  points  are  obtained  to  determine  the 
form  «f  the  curve. 

The  plotting  is  facilitated  by  the  use  of  polar  coordinate 
paper,  which  enables  us  to  plot  values  of  6  by  lines  drawn 
through  the  pole  and  values  of  p  by  circles  having  the  pole  as 
center.  The  tables  on  page  6  are  to  be  used  in  constructing  tables 
of  values  of  p  and  0. 

EXAMPLES 

1.  Plot  tlie  locus  of  the  equation 
(1)  /o  =  10  cos  (9. 

Solution.  The  calculation  is  made  by  assuming  values  for  6,  as  in  the 
table,  and  calculating  p,  making  use  of  the  natural  values  of  the  cosine 
given  in  Art.  4.   For  example,  if 

e  =  105°,  /3  =  10  cos  105°  =  10  cos  (180°-  75°)  =  -  10  cos  75°  =  -  2.6. 

90° 


p  =  lOcos0 

e 

P 

e 

P 

0 

10 

105° 

-   2.G 

15° 

9.7 

120° 

-   5 

30° 

8.7 

135° 

—    7 

45° 

7 

150° 

-   8.7 

G0° 

5 

1G5° 

-   0.7 

75° 

2.6 

180° 

-10 

90° 

0 

1 

The  complete  locus  is  found  in  this  example  without  going  beyond  180° 
for  9.   The  curve  is  a  circle  (Art.  50). 

Since  cos(— ^)  =  cos^  (29,  p.  3),  equation  (1)  may  be  written 
p  =  10  cos  (—  6) ,  that  is,  for  every  point  (p,  6)  on  tho  locus  there  is  also 


122 


NEW  ANALYTIC   GEOMETRY 


a  second  point  {p,  —  6)  on  the  locus.  Since  tliese  points  are  symmetrical 
with  respect  to  the  polar  axis,  we  have  the  result :  The  locus  of  (1)  is 
symmetrical  with  respect  to  the  polar  axis. 

2.  Draw  the  locus  of 
(2)  p^  =  a2  cos  2  0. 

Solution.   Before  plotting,  we  make  the  following  observations  : 

1.  Since  the  maximum  value  of  cos  2  ^  is  1,  the  maximum  value  of  p  is  a, 
and  the  curve  must  be  closed. 

2.  When  cos  2  ^  is  negative,  p  will  be  imaginary.    Now  cos  2  ^  is  nega- 
tive when  2  ^  is  an  angle  in  the  second  or  third  quadrant.    That  is,  when 

90°  <  2  6*  <  270°,  that  is,  45°  <  ^  <  135°, 

p  is  imaginary.  There  is  no  part  of  the  curve  between  the  45°  and  135° 
lines. 

3.  We  may  change  ^  to  —  ^  in  (2)  without  affecting  the  equation,  and 
hence  the  locus  is  symmetrical  with  respect  to  the  polar  axis. 

The  complete  curve  is  obtained  if  0  is  given  values  from  0°  to  45°,  as  in 
the  table. 


p2  =a2cos2e 

e 

20 

cos  2  6 

P 

0 

0 

1 

±rt 

15° 

30° 

.866 

±  .93  a 

30° 

60° 

.500 

±  .7  a 

45° 

90° 

0 

0 

The  complete  curve  results  by  plotting  these  points  and  the  points 
symmetrical  to  them  with  respect  to  the  polar  axis.  The  curve  is  called 
a.  lemniscate.    In  the  figure  a  is  taken  equal  to  9.5. 

3.  Discuss  and  plot  the  locus  of  the  equation 
iS)  p  =  a  sec2  ^  ft. 


POLAR  COORDINATES 


12S 


For  convenience  we  change  the  form  of  the  equation.   Using  (26),  p.  3, 

_       ^ 

Then  by  (41),  p.  4,  cos^l^  =  A  +  ^  cos^.    Hence  the  result : 


P  = 


2a 


1  +  COS( 


P 

=  2  -  (1  +  cos  9) 

e 

cos  e 

1  +  cos  e 

P 

e 

cos  0 

1  +  cos  e 

P 

0 

1 

2 

1 

105° 

-.259 

.741 

2.7 

15^ 

.966 

1.966 

1.02 

120° 

-.500 

.500 

4 

30° 

.866 

1.866 

1.07 

135° 

-.707 

.293 

6.7 

45° 

.707 

1.707 

1.2 

150° 

-.866 

.134 

14 

60° 

.500 

1.500 

1.3 

165° 

-.906 

.034 

50 

75° 

.259 

1.259 

1.6 

180° 

-1 

0 

CO 

90° 

0 

1 

2 

Solution.   Before  plotting,  we  note 

1.  The  curve  is  symmetrical  with  respect  to  the  polar  axis,  since  & 
may  be  replaced  by  —  0. 

2.  p  becomes  infinite 
when  1  +  cos^  =  0,  or 
cos  0  =  —I,  and  hence 
6  =  180°.  The  curve  re- 
cedes to  infinity  in  the 
direction  0  =  180°. 

3.  p  is  never  imaginary. 
On    account    of    1    the 

table  of  values  is  com- 
puted only  to  0  =180°,  and 
the  rest  of  the  curve  is  ob- 
tained from  the  .symmetry 
witli  re.spect  to  the  polar 
axis.  Take  a  =1.  The  locus 
is  a  parabola. 

Before  plotting  polar  equations,  the  student  should  establish 
such  simple  facts  as  result  froni  a  discussion,  as  illustrated  above. 


124 


NEW  AXALYTIC   (GEOMETRY 


PROBLEMS 
Plot  the  loci  of  the  following  equations : 


1. 

p  =  10. 

*  2. 

0  =  4.5^ 

-3. 

p  =  16  cos^. 

4. 

p  cos  6  —  Q. 

6. 

p  sin  ^  =  4. 

6. 

4 

1  -  cos  (9 

7. 

8 

^      2-cos(9 

8. 

8 

^       1  -  2  cos  ^ 

9. 

p  =  a  sin  ^. 

10. 

10 

^       1  +  tan  0 

11. 

p^  sin  2^  =  16. 

12. 

p"  cos  2  ^  =  a^. 

13. 

p  cos  0  =  a  sin^  ^. 

15.  p  =  a{l—  cos6). 

CARDIOID 


14.  p  =  a  sec  0  ±  b.   b  <  a. 

CONCHOID    OK    NICOMEDES 


16.  p-  =  a2  sin  2  0. 

TWO-LEAVEU    ROSE    LEMNISCATK 
F 

Q 

17.  p  =  b  —  a  cos^.   6  <  (X. 

LIMAgON 


POLAR  COORDINATES 


125 


18.  Plot  the  conchoid  (Problem  14)  for  h  =  a;  b  >  a. 

19.  Plot  the  lima^on  (Problem  17)  for  b  >  a. 

47.  The  student  should  acquire  skill  in  plotting  polar  equa- 
tions rapidly  when  a  rough  diagram  will  serve. 

For  example,  to  draw  the  locus  of 

(1)  p  =  a  sin  3^, 

we  proceed  as  follows  : 

Let  6  increase  from  0°.    Follow  the  variation  of  p  from  (1)  as  3  0 
tescribes  the  successive  quadrants. 


When  3 e  varies  from 

0°  to  90° 

90°  to  180° 

180°  to  270° 

270°  to  360° 

3G0°  to  450°  450°  to  540° 

then  0  varies  from 

0°  to  30° 

30°  to  60° 

60°  to  90 

90°  to  120° 

120°  to  150°  150°  to  180° 

anil  p  varies  from 

0  to  n 

a  toO 

0  to  -  rt 

-  rt  to  0 

0  to  a         a  to  0 

For  example,  when  3  &  varies  from  270°  to  360°,  that  is,  is  an  angle  in 
ihe  fourth  quadrant,  then  p  is  negative  and  increases  from  —  a  to  0. 

Now  draw  the  radial  lines 
corresponding  to  the  inter- 
vals of  d;  that  is,  0°,  30°,  60°, 
90°,  120°,, 150°,  180°. 

Noting  the  variation  of 
p,  we  sketch  the  curve  as 
follows : 

The  curve  starts  from  the 
pole  in  the  direction  0°, 
crosses  the  30°  line  perpen- 
dicularly at  p  =  a,  returns  to 
and  passes  through  the  pole 
on  the  60°  line,  crosses  the 
90°  line  produced  at  p  =  —  a, 
returns  to  and  passes  through 
the  pole  on  the  120°  line  (produced),  crosses  the  150°  line  at  /o  =  a,  and 
returns  to  the  pole  on  the  180°  line. 

This  gives  the  complete  locus.  The  pencil  point  has  moved  continu- 
ously without  abrupt  change  in  direction,  and  has  returned  to  the  original 
position  and  direction. 

The  curve  is  called  the  three-leaved  rose. 


126 


NEW  ANALYTIC   GEOMETRY 


PROBLEMS 
Draw  rapidly  the  locus  of  each  of  the  following  equations ; 
1.  p  =  acos3^.  3.  p  =  acos2^. 

3_  y 


THREE-LEAVED    ROSE 


FOUR-LEAVED    ROSE 


4.  /o  =  a  sin4( 


FOUR-LEAVED    ROSE 

10.  p  =  acos{0  +  45°). 

11.  p  =  a  sin  I  ^  +  ^|. 

12.  p  =  a  sin  1 0. 

13.  p  =  a  cos-- 


EIGHT-LEAVED    ROSE 


5.  p  =  a  cos  4^. 

6.  p  =  a  sin  5  0. 
1,  p  =  a  cos  5^. 

8.  p  =a{l  +  sin^) 

9.  p  =  a(l-^  cos (9) 


14.  p  =  a  sin  6  0 

15.  p  —  asin^^f 

16.  p  =  acos2^( 

17.  p  =:  a  sin^l  ( 

18.  p  =  a  cos^  \  { 


POLAR  COORDINATES 


127 


48.  Points  of  intersection.  By  a  method  analogous  to  that  used 
in  rectangular  coordinates  we  find  the  coordinates  of  the  points 
of  intersection  of  two  polar  curves  by  solving  their  equations 
simultaneously.  This  is  best  done  by  eliminating  p,  which  will 
give  rise  in  general  to  a  transcendental  equation  in  6  which 
can  be  solved  either  by  inspection  or  by  the  graphical  method 
employed  in  Art.  44. 

The  following  example  will  illustrate  the  method. 


EXAMPLE 

rind  the  points  of  intersection  of 

(1)  /D  =  l+COS(9, 

(2)  P  ^ 


2(1- cos  ^) 


Solution.    Eliminating /3, 
1  +  cos  (9 

1  —  C0S2  0 


COS  6  =  ± 


2(1—  COS  (9) 

h 

V2 


2 


.-.  e  =  ±  45°,  ±  135° 
Substituting  these  values  in  either  equation,  we  obtain  the  following 
four  points, 

1  +  -:^.    ±45°),    (1--^,   ±135°). 


2  /     \  2 

The  result  checks  in  the  figure.    The  locus  of  (1)  is  a  cardioid  ;  of  (2), 
a  parabola. 

PROBLEMS 

Find  the  points  of  intersection  of  the  following  pairs  of  curves  and 
check  by  drawing  the  figure  : 


r4/3cos6'  =  3, 
•  t2p  =  3. 

4  p  cos  ^  =  3, 
p  =  3  cos  6. 
'2p  =  S, 
/)  =  3  sin  0. 


4.  ^ 


fp  =V3, 
[p  =  2  sin  (9. 


r  /3  =  cos  0, 

'  \4/3  =  3sec(9. 
p  =  1  +  cos  0, 
2p  =  3. 


'^  2 

p  =  2. 

r  3  /o  =  4  cos  0, 

1   2/3COS2-  =  l. 


1128 


NEW  ANALYTIC   GEOMETRY 


10.  i 


=  sin  6/, 
=  cos  2  0. 
Am.  (J,  30°),  {i,  150P). 

p  =  1  +  cos  6, 
/)(!  +  cos  6')  =  1. 

A}is.  (1,  ±  90°). 

rp  =  2{l-sin^), 
■  \p(H-sin^)  =  L 

Ans.  (2  tV2,  ±  45°), 
(2tV2,  ±  135°). 

fp  =  4(1  +  cosi9), 
|p(l-cos^)  =  3. 
Ans.  (6,  ±  60°),  (2,  ±  120°). 


13.  i 


p  =  5—  2sinff, 
6 


L  l  +  sin^ 

C  /3  =  3  —  2  cos  ( 


14.  ^ 


[' 


3  +  2  cos  ^ 


15.  ■> 


(  p-  =  0  cos : 


IP 


^'  G  cos  ( 


16. 


f  p2  =  sin  2  e, 
\p  =  V2smd 


17.  -i 


f  p  =  cos  3  ( 


12 


cos^. 


18.  -; 


{2p 

(p=:0, 

IP  =  cos  8 


Let 


^  49.  Transformation  from  rectangular  to  polar  coordinates. 
OX  and  OF  be  the  axes  of  a  rectangular  system  of  coordinates, 
and  let  O  be  the  pole  and  OX  the  i)olar  axis  of  a  system  of 

r 


polar  coordinates.    Let  (.r,  ?/)  and  (p,  6)  be  respectively  the  rec- 
tangular and  polar  coordinates  of  any  point  P.    It  is  necessary 
to  distinguish  two  cases  according  as  p  is  positive  or  negative. 
When  p  is  positloe  (Fig.  1)  we  have,  by  definition, 

cos  p  =  -  ,     sui  Q  =  -  1 
P  P 

whatever  quadrant  P  is  in. 

Hence 

(1)  X  =  p  cos  6,     y  =  p  sin  6. 


POLAR  coordinatp:s  129 

When  p  is  negative  (Fig.  2)  we  consider  the  point  /*'  sym- 
metrical to  /'  with  res])ect  to  0,  whose  rectangular  and  polar 
coordinates  are  respectively  (—  .r,  —  y)  and  (—  p,  6).  The  radius 
vector  of  ]'',  —  p,  is  positive,  since  p  is  negative,  and  we  can 
therefore  use  equations  (1).    Hence  for  /-"' 

—  X  =  —  p  cos  9,     —  y  =  —  p  sin  0 ; 

and  hence  for  P 

X  =  p  cos  $,  y  =  p  sin  ^, 

as  before. 

Hence  we  have  the 

Theorem.  If  the  pole  coincides  icith  the  origin  and  the  polar 
axis  witli  the  positive  x-axis,  then 

\y  —  p  sin  B, 

ivhere  (.r,  ?/)  are  tlie  rcctangaiar  coordinates  and  (^p,  6)  the  polar 
coordinates  of  any  point. 

Equations  (I)  are  called  the  equations  of  transformation  from 
rectangular  to  polar  coordinates.  They  express  the  rectangular 
coordinates  of  any  point  in  terms  of  the  polar  coordinates  of 
that  point  and  enable  us  to  find  the  equation  of  a  curve  in  polar 
coordinates  when  its  equation  in  rectangular  coordinates  is 
known,  and  vice  versa. 

From  the  figures  we  also  have 


(2) 


'  X 

^  y  «  ^ 

s\nO  = 1     cos^ 


I 


±  V^T?  ±  ^^  +  y' 


These  equations  express  the  polar  coordinates  of  any  point  in 
terms  of  the  rectangular  co(')rdinates.  They  are  not  as  con- 
venient for  use  as  (T),  although  the  first  one  is  at  times  very 
convenient. 


130  NEW  ANALYTIC   GEOMETRY 

EXAMPLES 

1.  Find  the  equation  of  the  circle  x'^  +  y"  =  25  in  polar  coordinates. 
Solution.    From  the  first  equation  of  (2),  we  have  at  once  p^  =  25 ;  lience 

p  =  ±  5,  which  is  the  required  equation.    It  expresses  the  fact  that  the 
point  (p,  0)  Is  live  units  from  the  origin. 

2.  Find  the  equation  of  the  lemniscate  (Ex.  2,  p.  122)  p-  =  a^  cos  2  0  in 
rectangular  coordinates. 

Solution.    By  39,  p.  4,  since  cos  2^  =  cos"^^  —  sin^^, 

p-  —  a'^{cos"d—  sin-^). 
Substituting  from  (2), 

X-  +  y-  =  a-( ^  ■ 

.•.  (x-  +  y-)-  =  a"-^(x-  —  ?/2).    Ans. 
50.  Applications.    Straight  line  and  circle. 

Theorem.  The  general  equation  of  the  straight  line  in  jJolar 
coordinates  is 

(II)  p(AeosO  -{-  B  sin  6)  +  C  =  0, 
where  A,  B,  and  C  are  arbitrary  constants. 

Proof.  The  general  equation  of  the  line  in  rectangular  coordi- 
iiates  is  ^4  3.  ^  By -\-  C  =  0. 

By  substitution  from  (I)  we  obtain  (II).  q.e.d. 

Special  cases  of  (II)  are  pcos^=«,  psin^=6,  which  I'esult  respectively 
when  JS  =  0,  or  ^t  =  0  ;   that  is,  when  the  line  is  parallel  to  OY  or  OX. 

In  like  manner  we  obtain  from  (II), 
p.  93,  the 

Theorem.  The  general  equation  of  the 
circle  In  polar  coordinates  Is 

(III)  p-+  p(D  eosO  +  E  sin  &)  +  F=0, 
where  D,  E,  and  F  are  arhltrarij  constants. 

We  may  easily  show  further  that  if  the  pole  is  on  the  cir- 
cumference and  the  polar  axis  is  a  diameter,  the  equation  of 
the  circle  is  p  =  2rcos^, 

where  r  is  the  radius  of  the  circle. 


POLAR   COORDIXATES 


131 


For  if  the  center  lies  on  the  polar  axis,  or  a:"-axis,  E  =  0, 
and  if  the  circle  passes  through  the  pole,  or  origin,  F=  0.  The 
abscissa  of  the  center  equals  the  radius, 

and  hence  —  ■;j  =  >",  or  1)  =  —  2  r.  Substi- 
tuting these  values  of  D,  E,  and  /-'  in  (III) 
gives  p  —  2r  cos  ^  =  0. 

This  result  is  easily  seen  also  directly 
from  the  figure  on  page  130. 

Similarly,  if  the  circle  touches  the  polar  axis  at  the  pole,  the 
equation  is  p  =  2  r  sin  6. 

Theorem.     Tlte  h'mjth  I  of  the  Ime  Joining  tiro  jjolnts  l\{p.,  0) 
and  /'.,(po!  ^o)  '*'  ffu-en  hy 
(IV)  P  =  pi  +p^-2  p,p,  cos  (6,  -  e,) . 

Froof.    Let  the  rectangular  coordinates  of  P^  and  P^  be  re- 
spectively (x^,  7/j)  and  (a-.,,  y.,).    Then  by  (I),  p.  129, 
x^  =  p^  cos  6^,     :i\^  =  p,^  cos  ^.„ 
y^  =  p^  sin  0^,     ?/.,  =  p.,  sin  ^.,. 

But  /•^  =  (.,.^_^j^  +  (_y^_y./, 

and  hence     /-  =  (p^  cos  6^  —  p.,  cos  ^.,)-  +  (p^  sin  6^  —  p.^  sin  O.^y. 

Removing  parentheses  and  using  28  and  36,  p.  3,  Ave  o\>- 
tain  (IV).  O-E.D. 

Formula  (IV)  may  also  be  derived  directly  from  a  figure  by 
using  the  law  of  cosines  (44,  p.  4). 


rdinates  of  the  points  (s,  7  )    (—2,  —  j 


PROBLEMS 

1.  Find  the  polar  coordinates  of  the  points  (3,4),  (—4,  3),  (5,  —12),  (4,  5). 

2.  Find  the  rectangular  coon 

(3,   TT). 

3.  Transform  the  following  equations  into  polar  coordinates  and  plot 
their  loci: 

(a)  X  —  Sy  =  0.  Ans.    0  =  tau-^i. 

(b)  y^  +  5x  =  0.  Ans.   p  =  —  5  cot  0  cosec  9. 


132 


NEW    ANALY'lIC    (iKOAIKTRY 


(c)  x^  +  2/^  =  1().  Alls,  p  =  ±  4. 

^    (d)  x^  +  y'^  —  <ix  =  0.  Ans.  p  =  a  cos  6. 

,{c')2x(/  =  7.  Ans.  p^ am  2 6  =  1 . 

(f)  x^  —  y'^  —  ((-.  -.!)(«.  p^co?,2  6  =  a''^. 

(g)  X  cos  w  +  ?/  sin  w  —  p  =  0.  /I  n.s.  p  cos  {9  —  w)  —  p  =  d 

4.  Traiisfonii  equations  1  to  18.  p.  124,  into  rectangular  coordinates. 


LOCUS  PROBLEMS 

The  locus  should  be  drawn  in  each  case  (.see  the  figures  below). 

1.  Find  the  locus  of  a  point  such  that 

(a)  its  radius  vector  is  proportional  to  its  vectorial  angle. 

Ans.   The  spiral  of  Archimedes,  p  =  aO. 

Y 


p-y  -=  a-,    (p  >  0,\ 

LITUUS 


SriKAI.     OF    Al!(  Ul.MKDES 


HYFKHHOLIC    OH    KKl 
Sl'lliAI. 


l.tWAKITIlMIC    OK    E«}IIIAN- 
GII.AR    gPIKAL 


POLAR  COORDINATES 


133 


(b)  its  rarlius  vector  is  inversely  proportional  to  its  vectorial  angle. 

Ans.   The  hyperbolic  or  reciprocal  spiral,  pd  =  a. 

(c)  the  square  of  its  radius  vector  is  inversely  proportional  to  its 
vectorial  aniile.  Ans.    The  lituus,  p^O  =  a^. 

(d)  the  l(»i;aritlnn  of  its  radius  vector  is  proportional  to  its  vectorial 
angle.  Ans.   The  logarithmic  spiral,  log p  =  a^. 

Theorem  on  the  logarithmic  spiral.  When  two  points,  Pj  and  P,,  have 
been  plotted  on  a  logarithmic  spiral,  points  between  them  on  the  locus 
may  be  constructed  geometrically  by  the  following  theorem  : 

If  the  angle  P^OP-^  is  bisected,  and  if  on  this  bisector  OP3  is  laid  off  equal 
to  a  mean  proportional  between  OP.^  and  0P„,  then  Pg  is  on  the  locus. 

Proof.    By  hypothesis,  since  Pj  and  P^  are  on 
the  curve  log  p  =  aO, 
(1)  log  /)j  =  a^j    and    log/?.,  =  aO^. 

Adding  and  dividing  by  2, 

h  +  ^2^ 


(2) 


log 


2   l0gpi+   I  logP2 


/^1  + 


(14  and  17,  p.  1). 


If  Pg  is  (pj,  ^3),  then,  by  construction. 


=  (^3  -  ^1,  or  (93 


+ 


: ,  and  p^  =  V, 


P\P-2 


Hence,  by  (2),  logpg  =  ad^,  and  P^  is  also  011  the  locus. 


R.{po,9o) 


PiiPudi) 


Q.E.D. 


PROBLEMS  FOR  mDIVIDUAL   STUDY 

Plot  carefully  the  following  loci : 
1.  p  =  a  sin  ^  +  bsecO. 


2.  ip--)    =  a2cos2^. 

3.  p  =  a  (cos  2  (9  +  sin  2^). 
i.  p  =  a  cos  2  0  -\-  -  sec  0. 

5.  p  =  a  sill  2ff+-  sec  0. 

2 

6.  p  =  a  cos 20  +  b cos 0. 

7.  p  =  a  sin 20  +  b cos 0. 

8.  p  -  a  cos2  0  +  b{ii\n0  +  1). 

9.  p  =  a  cos  3^  —  6  cos  0. 


10. 

p  =  cos  3  ^  +  cos  0  +  1 

11. 

p  =  cos  3  ^  +  cos  2  0. 

12. 

p  =  cos  3  6*  —  sin  2^. 

13. 

.   .0 
p  =  asin3-. 

14. 

p  =  a  cos^  -  . 
'^                2 

15. 

p'^cos0  =  a2sin3^. 

16. 

0      2cos^ 
^   ~cos2^        ■ 

17. 

2  cos  2  (9 
P^  = ^-^  +  1- 

cos  (9  +  2 


CHAPTER  VIII 


FUNCTIONS  AND  GRAPHS 


51.  Functions.  In  many  practical  problems  two  variables 
are  involved  in  such  a  manner  that  the  value  of  one  depends 
upon  the  value  of  the  other.  For  example,  given  a  large  num- 
ber of  letters,  the  postage  and  the  weight  are  variables,  and 
the  amount  of  the  postage  depends  upon  the  weight.  Again,  the 
premium  of  a  life-insurance  policy  depends  upon  the  age  of  the 
applicant.    Many  other  examples  will  occur  to  the  student. 

This  relation  between  two  variables  is  made  precise  by  the 
definition  : 

A  variable  is  said  to  he  a  function  of  a  second  variable  ichen 
its  value  depends  upon  the  value  of  the  latter  and  is  determined 
when  a  definite  value  is  assnvied  for  the  second  variable. 

Thus  the  jyostage  is  determined  when  a  definite  weight  is  as- 
sumed ;  the  premium  is  determined  when  a  definite  age  is  assumed. 

Consider  another  example : 

Draw  a  circle  of  diameter  5  in.  An 
indefinite  number  of  rectangles  may 
be  inscribed  within  this  circle.  But 
the  student  will  notice  that  the  entire 
rectawjle  is  determined  as  soon  as  a 
side  is  drawn.  Hence  the  area  of  the 
rectangle  is  ^function   of  its  side. 

Let  us  now  find  the  equation  ex- 
pressing the  relation  between  a  side  and  the  area  of  the  rectangle. 

Draw  any  one  of  the  rectangles  and  denote  the  length  of  its 

base  by   x  in.     Then  by   drawing   a   diagonal   (which  is,  of 

134 


FUNCTIONS  AND  GRAPHS 


135 


x'f. 


course,  a  diameter  of  the  circle),  the  altitude  is  found  to  be 
equal  to  (25  —  x^y.  Hence  if  A  denotes  the  area  in  square 
inches,  we  have 
(1)  A  =  x(25 

This  equation  gives  the  functional  relation  between  the  func- 
tion .4  and  the  variable  x.  From  it  we  are  enabled  to  calculate 
the  value  of  the  function  A  corresponding  to  any  value  of  the 
variable  x.    For  example  : 

if  x  =  l  in.,     .1  =  (24)^  =  4.9  sq.  in. ; 

if  ic  =  3in.,     yl=12sq.  in. ; 

if  x  =  A  in.,     .4=12  sq.  in. ;  etc. 

To  obtain  a  representation  of  the  equation  (1)  for  all  vraues 
of  X,  we  draw  a  graph  of  the  equation.  This  we  do  by  draw- 
ing rectangular  axes  and  plotting 

the.  values  of  the  variable  (x)  as  abscissas, 
the  values  of  the  function  (.4)  as  ordinates. 

Any  functional  relation  may  be  graphed  in  this  way.  We 
must,  however,  first  discuss  the  equation  (1). 

The  values  of  x  and  A  are  positive  from  the  nature  of  the 
problem. 

The  values  of  x  range  from  zero  to 
5,  inclusive. 

The  student  should  now  choose  a 
suitable  scale  on  each  axis  and  draw 
the  graph.  In  this  case,  unit  length 
on  the  axis  of  abscissas  represents 
1  in.,  and  unit  length  on  the  axis  of 
ordinates  represents  1  sq.  in.  These 
two  unit  lengths  need  not  be  the  same. 

What  do  we  learn  from  the  graph  ? 

1.  If  carefully  drawn,  we  may  measure  from  the  graph  the 
area  of  the  inscribed  rectangle  corresponding  to  any  side  we 
choose  to  assume. 


136  NEW  ANALYTIC  GEOMETRY 

'  2.  There  is  one  horizontal  tangent.  The  ordinate  at  its 
point  of  contact  is  greater  than  any  other  ordinate.  Hence 
this  discov^ery  :  One  of  the  inscribed  rectangles  is  greater  in  area 
than  any  of  the  otliers\  that  is,  there  is  a  maximum  rectangle. 
In  other  words,  the  function  defined  by  equation  (1)  has  a 
maximum  value. 

Careful  measurement  will  give  for  the  base  of  the  maximum 
rectangle,  x  =  3.5,  and  for  the  area,  A  =  12.5.  These  results, 
as  may  be  shown  by  the  methods  of  the  differential  calculus, 
are,  in  fact,  correct  to  one  place  of  decimals.  The  maximum 
rectangle  is  a  square ;  that  is,  of  all  rectangles  inscribed  in  a 
given  circle,  the  square  has  the  greatest  area. 

The  fact  that  a  maximum  rectangle  exists  can  be  seen  in 
advance  by  reasoning  thus  :  Let  the  base  x  increase  from  zero 
to  5  in.  The  area  A  will  then  begin  with  the  value  zero  and 
return  to  zero.  Since  .4  is  always  positive,  the  graph  must 
have  a  "  highest  point."  Hence  there  is  a  maximum  value  of 
A,  and  therefore  a  maximum  rectangle. 

Take  one  more  example :  A  wooden  box,  open  at  the  top,  is 
to  be  built  to  contain  108  cu.  ft.  The  base  must  be  square. 
This  is  the  only  condition.  It  is  evi- 
dent that  under  this  condition  any 
number  of  such  boxes  may  be  built, 
and  that  the  number  of  square  feet 
of  lumber  used  will  vary  accordingly. 
If,  however,  we  choose  any  length  for 
a  side  of  the  square  base,  only  one 
box  with  this  dimension  can  be  built, 

and  the  material  used  is  determined.    Hence  the  material  used 
is  a  function  of  a  side  of  the  square  base. 

Let  us  now  find  the  functional  relation  between  the  number 
of  square  feet  of  lumber  necessary  and  the  length  of  one  side 
of  the  square  base  measured  in  feet. 

Consider  any  one  box. 


A     / 

) 

FUNCTIONS  AND  GRAPHS 


137 


Let 


and 

Then 
and 

Hence 


M  =  amount  of  lumber  in  square  feet, 
X  =  length  of  side  of  the  square  base  in  feet, 
]i  =  height  of  the  box  in  feet, 
area  of  Vjase  =  x^  sq.  ft., 
area  of  sides  =  4  hx  sq.  ft. 
M=  x^  -\-  A  Jix. 

But  a  relation  exists  between  A  and  a-,  for  the  value  of  M 
must  depend  upon  the  value  of  x  alone.  In  fact,  the  volume 
equals  108  cu.  ft. 

Hence  hx"^ 

Therefore 

432 
x 


(2) 


108,    and    It  = — —• 


M=  x^  + 


This  equation  enables  us  to  calculate  the  number  of  square 
feet  of  lumber  in  any  box  with  a  given  square  base  which  has 
a  capacity  of  108  cu.  ft.    The  calculation  is  given  in  the  table  : 


X   0  1  1 

2 

4    5 

G 

7 

8 

20 

etc. 

feet 

M   00  i  433 

1 

220 

153 

124  111 

108 

111 

118 

421 

etc. 

sq.  ft. 

Thus,  if  X  =  1  ft.,     M  -  433  sq.  ft. ; 

if  a;  =  4  ft.,      .1/ =  124  sq.  ft. ; 

if  a;  =  8  ft.,     .1/  =  118  sq.  ft. ;  etc. 

The  student  should  now  graph  equation  (2),  choosing  units 
thus  : 

unit  length  on  the  axis  of  abscissas  represents  1  ft. ; 
unit  length  on  the  axis  of  ordinates  represents  1  sq.  ft. 

We  must,  however,  choose  a  very  small  unit  ordinate,  since  the 
values  of  .1/  are  large. 

A  preliminary  discussion  of  (2)  shows  that  x  may  have  any 
value  (ix)sitive). 


138 


NEW  ANALYTIC   GEOMETRY 


M 


1    2    3 


4    5    6 
Feet 


7    8    9   lOx 


What  do  we  learn  from  the  graph? 

1.  If  carefully  drawn,  we  may  measure  from  the  graph  the 
number  of  square  feet  of  lumber  in  any  box  which  contains 
108  cu.  ft.  and  has  a  square  base. 

2.  There  is  one  horizontal 
tangent.  The  ordinate  at  its 
point  of  contact  is  less  than 
any  other  ordinate.  Hence  this 
discovery:  One  of  the  boxes  takes 
less  himher  than  any  other ;  that 
is,  M  has  a  minimum  value.  This 
point  on  the  graph  can  be  deter- 
mined exactly  by  calculus,  but 
fiareful  measurement  will  in  this 

case  give  the  correct  values,  namely,  a?  =  6,  M  =  108.  That  is, 
the  construction  will  take  the  least  lumber  (108  sq.  ft.)  if  the 
base  is  6  ft.  square. 

The  fact  that  a  least  value  of  M  must  exist  is  seen  thus. 
Let  the  base  increase  from  a  very  small  square  to  a  very  large 
one.  In  the  former  case  the  height  must  be  very  great,  and 
hence  the  amount  of  lumber  will  be  large.  In  the  latter  case, 
while  the  height  is  small,  the  base  will  take  a  great  deal  of 
lumber.  Hence  M  varies  from  a  large  value  to  another  large 
value,  and  the  graph  must  have  a  "  lowest  point." 

In  the  following  problems  the  student  will  work  out  the 
functional  relation,  draw  the  graph,  and  state  any  conclusions 
to  be  drawn  from  the  figure.  Care  should  be  exercised  in  the 
selection  of  suitable  scales  on  the  axes,  especially  in  the  scale 
adopted  for  plotting  values  of  the  function  (compare  p.  137). 
The  graph  should  be  neither  very  flat  nor  very  steep.  To 
avoid  the  latter  w^e  may  select  a  large  miit  of  length  for  the 
variable.  The  plot  should  be  accurate  and  the  maximum  and 
minimum  values  of  the  function  should  be  measured  and  calcu- 
lated, additional  values  of  the  variable  being  used,  if  necessary. 


FUNCTIONS  AND  GRAPHS  139 

PROBLEMS 

1.  Rectangles  are  inscribed  in  a  circle  of  radius  2  in.  Plot  the 
perimeter  P  of  the  rectangles  as  a  function  of  the  breadth  x. 

Ans.    P  =  2x  +  2{l6-x^)i. 

2.  Right  triangles  are  constructed  on  a  line  of  length  5  in.  as  hypote- 
nuse. Plot  (a)  the  area  A  and  (b)  the  perimeter  P  as  a  function  of  the 
length  X  of  one  leg. 

Ans.    (a)  A=ix{2P>-  x^)^  ;  (b)  P  =  x  +  5  +  (25  -  x'^)^. 

3.  Right  cylinders*  are  inscribed  in  a  sphere  of  radius  r.  Plot  as  func- 
tions of  the  altitude  x  of  the  cylinder,  (a)  the  volume  V  of  the  cylinder, 
(b)  the  curved  surface  S. 

Ans.  (a)  V=  —  (4  r2x  -  x^) ;  (b)  ,S  =  irx  (4  r^  -  x'^)^. 

4.  Right  cones*  are  inscribed  in  a  sphere  of  radius  r.  Plot  as  func- 
tions of  the  altitude  x  of  the  cone,  (a)  the  volume  V  of  the  cone,  (b)  the 

curved  surface  S.  _  , 

Ans.    (a)  F  =  -  (2  rx^  -  x^) ;  (b)  S  =  tt  (4  r^-x^  -  2  rx^)^. 
o 

5.  Right  cylinders  are  inscribed  in  a  given  right  cone.  If  the  height 
of  the  cone  is  h  and  the  radius  of  the  base  r,  plot  (a)  the  volume  V  of 
the  cylinder,  (b)  the  curved  surface  S,  (c)  the  entire  surface  T,  as 
functions  of  the  altitude  x  of  the  cylinder. 

{h-x); 


(a) 

F  = 

7rr2x 

{h- 

-X)2;   (b)   S  = 

2Trrx 
h 

(c) 

T  = 

2Trr 

{h- 

-  x)  [rh  +  {h  - 

-r)x] 

6.  Right  cones  are  circumscribed  about  a  sphere  of  radius  r.  Plot  as 
a  function  of  the  altitude  x  of  the  cylinder,  the  volume  V  of  the  cone. 

T^x'^ 

Ans.    F  =  i  TT 

^     X  -  2  r 

7.  Right  cones  are  constructed  with  a  given  slant  height  L.  Plot  as 
functions  of  the  altitude  x  of  the  cone,  (a)  the  volume  V  of  the  cone, 
(b)  the  curved  surface  S,  (c)  the  entire  surface  T. 

Ans.    (a)  F  =  i  TT  (L2x  -  x^) ;  (b)  S  =  tt Z  {U  -  x^)*. 

8.  A  conical  tent  is  to  be  constructed  of  given  volume  F.  Plot  the 
amount  A  of  canvas  required  as  a  function  of  the  radius  x  of  the  base. 

,       (7r2x6  +  9  F2)^ 

J.TIS.     A=- —■ 

X 

*Use  formulas  .")-9,  p.  1. 


140  NEW  ANALYTIC   GEOMP:TPvY 

9.  A  cylindrical  tin  can  is  to  be  constructed  of  given  volume  V.    Plot 

the  amount  A  of  tin  required  as  a  function  of  the  radius  x  of  the  can. 

.       o      o      2  V 
Ann.    A  =  2  wx-  -\ 

X 

10.  An  open  box  is  to  be  made  from  a  sheet  of  pasteboard  12  in. 
square  by  cutting  equal  squares  from  the  four  corners  and  bending  up 
the  sides.  Plot  the  volume  V  as  a  function  of  the  side  x  of  the  square 
cut  out.  Ans.    T"  =  X  (12  —  2  x)"^. 

11.  The  strength  of  a  rectangular  beam  is  proportional  to  the  product 
of  the  cross  section  by  the  square  of  the  depth.  Plot  the  strength  S  as  a 
function  of  the  depth  x  for  beams  which  are  cut  from  a  log  12  in.  in 
diameter.  Ans.    S  =  kx3{lU  -  x^)i. 

12.  A  rectangular  stockade  is  to  be  built  to  contain  an  area  of  1000 

sq.   yd.  A  stone  wall  already  constructed  is  available  for  one  of  the 

sides.    Plot  the  length  L  of  the  wall  to  be  built  as  a  function  of  the  length 

x  of  the  side  of  the  rectangle  parallel  to  the  wall.       ^    „     t       2000 

°      ^  Ans.    L  = 1-  x. 

x 

13.  A  tower  is  100  ft.  high.  Plot  the  angle  y  subtended  by  the  tower 
at  a  point  on  the  ground  as  a  function  of  the  distance  x  from  the  foot  of 

X 

14.  A  tower  .5.5  ft.  high  is  surmounted  by  a  statue  10  ft.  high.  If  an 
observer's  eyes  are  5  ft.  above  the  ground,  plot  the  angle  y  subtended  by 
the  statue  as  a  function  of  the  observer's  distance  x  from  the  tower. 

,60  ,50 

Ans.    //=tan-i tan-i  — 

X  X 

15.  A  line  is  drawn  through  a  fixed  point  (a,  b).    Plot  as  a  function  of 

the  intercept  on  XX'  (=  x)  of  the  line,  the  area  A  of  the  triangle  formed 

with  the  coordinate  axes.  ^„„      a  bx^ 

Ans.   A  = 


2{x  —  «) 

16.  A  ship  is  41  mi.  due  north  of  a  second  ship.  The  first  sails  south 
at  the  rate  of  8  mi.  an  hour,  the  second  east  at  the  rate  of  10  mi.  an  hour. 
Plot  their  distance  d  apart  as  a  function  of  the  time  t  which  has  elapsed 
since  they  were  in  the  position  given.  ^^^^    ^  ^  ^^jy^a  _  gsei  +  I681)i 

17.  Plot  the  distance  e  from  the  point  (4,  0)  to  the  points  (x,  y)  on  the 
parabola  y-  =  4x.  Ans.    e  =  (x^  —  4x  +  16)^. 

18.  A  gutter  is  to  be  constructed  whose  cross  section  is  a  broken  line 
made  up  of  three  pieces,  each  4  in.  long,  the  middle  piece  being  horizon- 
tal, and  the  two  sides  being  equally  inclined,    (a)  Plot  the  area  A  of 


FUNCTIONS  AND  GRAPHS  141 

a  cross  section  of  the  gutter  as  a  function  of  the  width  x  of  the  gutter 
across  the  top.  (b)  Plot  the  area  J.  as  a  function  of  the  angle  of  incli- 
nation of  the  sides  to  the  horizontal. 

Ans.    (a)  A  =\{x+  4)  (48  +  8x  -  x2)i;  (b)^  =  8(sin  2^  +  2  .sin^). 

19.  A  Norman  window  consists  of  a  rectangle  surmounted  by  a  semi- 
circle. Given  the  perimeter  P,  plot  the  area  J.  as  a  function  of  the  width  x. 

Ans.    A  =  -£P x2 x2. 

2  2  8 

20.  A  person  in  a  boat  9  mi.  from  the  nearest  point  of  the  beach 

wishes  to  reach  a  place  15  mi.  from  that  point  along  the  shore.    He  can 

row  at  the  rate  of  4  mi.  an  hour  and  walk  at  the  rate  of  5  mi.  an  hour. 

The  time  it  takes  him  to  reach  his  destination  depends  on  the  place  at 

which  he  lands.    Plot  the  time  as  a  function  of  the  distance  x  of  his 

landing  place  from  the  nearest  point  on  the  beach.      /— 

°  ^  .         r^.  V  81  -f-  x^      15  —  X 

Ans.    Time  = 1 

4  5 

21.  The  illumination  of  a  plane  surface  by  a  luminous  point  varies 
directly  as  the  cosine  of  the  angle  of  incidence,  and  inversely  as  the 
square  of  the  distance  from  the  surface.  Plot  the  illumination  Z  at  a 
point  on  the  floor  10  ft.  from  the  wall  as  a  function  of  the  height  x  of  a 
gas  burner  on  the  wall.  j^^g     j  _  ^^ 

(100  +  ^2)2 

22.  A  Gothic  window  has  the  shape  of  an  equilateral  triangle  mounted 
on  a  rectangle.  The  base  of  the  triangle  is  a  chord  of  the  window.  The 
total  length  of  the  frame  of  the  window  is  constant.  Express,  plot,  and 
discuss  the  area  of  the  window  as  a  function  of  the  width. 

23.  A  printed  page  is  to  contain  24  sq.  in.  of  printed  matter.  The  top 
and  bottom  margins  are  each  1^  in.,  the  side  margins  1  in.  each.  Express, 
plot,  and  discuss  the  area  of  the  page  as  a  function  of  the  width. 

24.  A  manufacturer  has  96  sq.  ft.  of  lumber  with  which  to  make  a 
box  with  a  square  base  and  a  top.  Express,  plot,  and  discuss  the  contents 
of  the  box  as  a  function  of  the  side  of  the  base. 

25.  (a)  Isosceles  triangles  of  the  same  perimeter,  12  in.,  are  cut  out  of 
rubber.  Express,  plot,  and  discuss  the  area  as  a  function  of  the  base. 
(b)  Isosceles  triangles  of  the  same  area,  10  sq.  in.,  are  cut  out  of  rubber. 
Express,  plot,  and  discuss  the  perimeter  as  a  function  of  the  base. 

26.  Small  cylindrical  boxes  are  made  each  with  a  cover  whose  breadth 
and  height  are  equal.  The  cover  slips  on  tight.  Each  box  is  to  hold 
TT  cu.  in.  Express,  plot,  and  di.scuss  the  amount  of  material  used  as  a 
function  of  the  length  of  the  box. 


142  NEW  ANALYTIC  GEOMETRY 

27.  A  circular  filter  paper  has  a  diameter  of  11  in.  It  is  folded  into  a 
conical  shape.  Express  the  volume  of  the  cone  as  a  function  of  the  angle 
of  the  sector  folded  over.    Plot  and  discuss  this  function. 

28.  Two  sources  of  heat  are  at  the  points  A  and  B.  Remembering  that 
the  intensity  of  heat  at  a  point  varies  inversely  as  the  square  of  the  distance 
from  the  source,  express  the  intensity  of  heat  at  any  point  betvpeen  A 
and  B  as  a  function  of  its  distance  from  A.  Plot  and  discuss  this  function. 

29.  A  submarine  telegraph  cable  consists  of  a  central  circular  part, 
called  the  core,  surrounded  by  a  ring.  If  x  denotes  the  ratio  of  the  radius 
of  the  core  to  the  thickness  of  the  ring,  it  is  known  that  the  speed  of 

signaling  varies  as  x^  log-.    Plot  and  discuss  this  function. 

X 

30.  A  wall  10  ft.  high  surrounds  a  square  house  which  is  1.5  ft.  from 
the  wall.  Express  the  length  of  a  ladder  placed  without  the  wall,  resting 
upon  it  and  just  reaching  the  house,  as  a  function  either  of  the  distance 
of  the  foot  of  the  ladder  from  the  wall,  or  of  the  inclination  of  the  ladder 
to  the  horizontal.    Plot  and  discuss  this  function. 

31.  The  volume  of  a  right  prism  having  an  equilateral  triangular  base 
is  2.  Express  its  total  surface  as  a  function  of  the  edge  of  the  base. 
Plot  and  discuss. 

32.  A  letter  Y  .stands  a  ft.  high  and  measures  b  ft.  across  the  top. 
Express  the  total  length  of  the  leg  and  two  arms  as  a  function  of  the 
length  of  the  leg.    Plot  and  discuss. 

33.  The  sum  of  the  perimeters  of  a  square  and  a  circle  is  constant. 
Express  their  combined  areas  as  a  function  of  the  radius  of  the  circle. 
Plot  and  discuss. 

34.  A  water  tank  is  to  be  constructed  with  a  square  base  and  open  top, 
and  is  to  hold  64  cu.  yd.  The  cost  of  the  sides  is  |1  a  square  yard,  and 
of  the  bottom  |2  a  square  yard.   Plot  and  discuss  the  cost. 

35.  A  rectangular  tract  of  land  is  to  be  bought  for  the  purpose  of  lay- 
ing out  a  quarter-mile  track  with  straightaway  sides  and  semicircular 
ends.  In  addition  a  strip  3-5  yd.  wide  along  each  straightaway  is  to  be 
bought  for  grand  stands,  training  quarters,  etc.  If  the  land  costs  |200 
an  acre,  plot  and  discuss  the  cost  of  the  land  required. 

36.  A  cylindrical  steam  boiler  is  to  be  constructed  having  a  capacity 
of  1000  cu.  ft.  The  material  for  the  side  costs  |2  a  square  foot,  and  for 
the  ends  |.3  a  square  foot.    Plot  and  discuss  the  cost. 

37.  In  the  corner  of  a  field  bounded  by  two  perpendicular  roads  a 
spring  is  situated  6  rd.  from  one  road  and  8  rd.  from  the  other.    How 


FUNCTIONS  AND  GRAPHS  143 

-should  a  straight  road  be  run  by  this  spring  and  across  the  corner  so  as 
to  cut  off  as  little  of  the  field  as  ijossible  ? 

Ans.    12  and  16  rd.  from  the  corner. 
38.  When  the  resistance  of  air  is  taken  into  account,  the  inclination  of 
a  pendulum  to  the  vertical  is  given  by  the  formula 

0  =  ae-^' cos  (nt  +  e). 
Plot  ^  as  a  function  of  the  time  t. 

52.  Notation  of  functions.  The  symbol  f  (x)  is  used  to  de- 
note a  function  of  x,  and  is  read  "/  of  x."  In  order  to  distin- 
guish between  different  functions,  the  prefixed  letter  is  changed, 
as  F(x),  <f>  (x)  (read  "^7u*  of  x  "),f(x),  etc. 

During  any  investigation  the  same  functional  symbol  always 
indicates  the  same  law  of  dependence  of  the  function  upon  the 
variable.  In  the  simpler  cases  this  law  takes  the  form  of  a 
series  of  analytical  operations  upon  that  variable.  Hence,  in 
such  a  case,  the  same  functional  symbol  will  indicate  the  same 
operations  or  series  of  operations,  even  though  applied  to 
different  quantities.   Thus,  if 

f{x)=x^-9x  +  U, 

then  /(y)  =2/-9  }/-\-  14. 

Also  f(a)=a''-9a  +  14:, 

f{b  ^l)=(b  +  iy-9{b  +  l)  +  14:  =  b^-7b  +  6, 
/(0)=02-90  +  14  =  14, 
/(-l)  =  (-lf-9(-l)  +  14=24, 
/(7)  =  72  -  9  ■  7  +  14  =  0,  etc. 

PROBLEMS 

1.  Given  0(x)  =  log^a;.  Find  ^(2),  ^(1),  0(5),  0(a-l),  4>{b'^), 
<f>{x+  1),  <t>(Vx). 

2.  Given  0(x)  =  e^^.    Find  0(0),  0(1),  0(-  1),  0(2?/),  <p{-x). 

3.  Given/(x)=sin2x.  Find /(^),/(^), /(  -  rr), /(- a:), /(tt- x), 
/(i7r-^),/(i7r +  ii).  ^"^^       ^*^ 

4.  Given  0  (x)  =  cos  x.    Prove 

«(x)  +  «W  =  2«(i±i!),(^-^). 


CHAPTER  IX 


TRANSFORMATION  OF  COORDINATES 


53.  When  we  are  at  liberty  to  choose  the  axes  as  we  please 
we  generally  choose  them  so  that  our  results  shall  have  the 
simplest  possible  form.  When  the  axes  are  given,  it  is  impor- 
tant to  be  able  to  find  the  equation  of  a  given  curve  referred 
to  some  other  axes.  The  operation  of  changing  from  one  pair  of 
axes  to  a  second  pair  is  known  as  a  transformation  of  coordinates. 
We  regard  the  axes  as  moved  from  their  given  position  to  a  new 
position  and  we  seek  formulas  which  express  the  old  coordi- 
nates in  terms  of  the  ne\\j  coordinates. 

54.  Translation  of  the  axes.  If  the  axes  be  moved  from  a  first 
position  OX  and  OF  to  a  second  position  O'X'  and  O'Y'  such  that 
O'.Y'and  O'F'are  respectively  par- 
allel to  OX  and  0  Y,  then  the  axes 
are  said  to  be  translated  from  the 
first  to  the  second  position. 

Let  the  new  origin  be  0'  (Ji,  kj 
and  let  the  coordinates  of  any 
point  P  before  and  after  the 
translation  be  respectively  (x,  y) 
and  {x\  y').    Then,  in  the  figure, 

ai=//,      OM  =  x,      0'M'  =  x', 

OB  =  k,     MP  =  y,     M'P  =  y\ 

Projecting  OP  and  00' P  on  OX,  we  obtain  (Art.  31) 

OM  =  OA  +  O'M' ; 

.' .  X  =  x'  -\-  h. 
Similarly,  y  =  //'  +  k. 

1-J4 


TRANSFORMATION  OF  COORDINATES 


145 


Hence  the 

Theorem.  If  the  axes  he  translated  to  a  new  origin  (h,  k'),  and 
if  (x,  y)  and  (x',  y')  are  respectixiely  the  coordinates  of  any  point 
P  before  and  after  the  translation,  then 


(I) 


(x=x^  -if-h, 


iy  =  y'  +  k. 

Equations  (I)  are  called  the  equations  for  translating  the  axes. 
To  find  the  equation  of  a  curve  referred  to  the  new  axes  when 
its  equation  referred  to  the  old  axes  is  given,  substitute  in  the 
given  equation  the  values  of  r  and  y  given  by  (I)  and  reduce. 

EXAMPLE 

Transform  the  equation 

X-  +  y~  —  6x  +  4y  —  12  =  0 
when  the  axes  are  translated  to  the  new  origin  (3,  —  2). 

Solution.  Here  h  =  S  and  k  =  —  2, 
so  equations  (I)  become 

X  =  x'  +  3,     y  =  y'-2. 
Substituting  in  the  given  equation, 
we  obtain 

(x'  +  3)2+  (y'_2)2-r)(x'  +  3) 
+  4(//-2)-12  =  0, 
or,  reducing,  x"-  +  y'-  =  25. 

This  result  could  easily  be  fore- 
seen. For  the  locus  of  the  given 
equation  is  a  circle  whose  center  is 
(3,  —  2)  and  whose  radius  is  5.  When 
the  origin  is  translated  to  the  center  the  equation  of  the  circle  must  necessa- 
rily have  the  form  obtained. 

PROBLEMS 

1.  Find  the  new  coordinates  of  the  points  (3,  —  5)  and  (—  4,  2)  when 
the  axes  are  translated  to  the  new  origin  (3,  6). 

2.  Transform  the  following  equations  when  the  axes  are  translated  to 
the  new  origin  indicated  and  plot  both  pairs  of  axes  and  the  curve  : 

(a)  3x  — 4?/ =  0,  (2,  0).  Ans.    3x'— 42/'  =  0. 

(b)  x2  + y/2_43._2j/=:0,  (2,  1).  Ana.    x'2  +  ?/'2  =  .5. 


r^ 

< 

\" 

~ 

^ 

^^ 

\ 

/ 

\ 

/ 

\ 

/ 

0 

\ 

X 

0' 

(3, 

-2) 

X 

y 

) 

\ 

/ 

\ 

/ 

■v. 

' 

^ 

^ 

146 


NEW  ANALYTIC   GEOMETRY 


(c)  ?/2_6x  + 9  =  0,(1,0).  Ans.   ?/2  =  6x'. 

(d)  x2  +  ?/2_i  =  0,  (- 3,  —  2).  Ans.   x'^  +  y'^-Qx'-4y'+12=0. 

(e)y^-2kx  +  k^  =  0,(-,o\.  Ans.    y"^  =  2kx\ 

(f)  x2-4?/2  +  8a;  +  24?/-20  =  0,  (-4,3).    Ans.   x'2-4y'2_o. 

3.  Derive  equations  (I)  if  0'  is  in  (a)  tlie  second  quadrant ;  (b)  the 
tliird  quadrant ;  (c)  the  fourth  quadrant. 

55.  Rotation  of  the  axes.  Let  the  axes  OX  and  OF  be  rotated 
about  0  through  an  angle  $  to  the  positions  OX'  and  OY'. 
The  equations  giving  the 
coordinates  of  any  point  y 
referred  to  OX  and  OY  in 
terms  of  its  coordinates  re- 
ferred to  OX'  and  OY'  are 
called  the  equations  for  ro- 
tating the  axes. 

Theorem.  The  equations  for  rotating  the  axes  through  an 
angle  6  are 

fX=x'cos9~y'  sin 6, 

^     ^  \y  =  x' sind  +  y' cosd. 

Proof.  Let  P  be  any  point  whose  old  and  new  coordinates 
are  respectively  (x,  ij)  and  (x',  y').  Draw  OP,  and  draw  PM' 
perpendicular  to  OX'.    Project  OP  and  OM'P  on  OX. 

The  projection  of  OP  on  OX  =  x.  (Art.  31) 

The  projection  of  OM'  on  OX  =  cc'cos^.  (Art.  31) 

The  projection  of  M'P  on  OX  =  y'  cos  (^  +  o)         (Art.  31) 

=  -7/' sine.  (By31,  p.  3) 

But  by  Art.  31, 

projection  of  OP  =  projection  of  OM'  -f  projection  of  M'P. 

.'.  x  =  ic'cos^  —  y' &\i\d. 


TRANSFORMATION  OF  COORDINATES  147 

In  like  manner,  projecting  OP  and  OM'P  on  OY,  we  obtain 

2/ =  ic' cos  /  —  —  ^  j  +  7/' cos  ^ 

=  cc'sin^ -(- y'cos^.  Q.E.D. 

If  the  equation  of  a  curve  in  x  and  y  is  given,  we  substitute 

from  (II)  in  order  to  find  the  equation  of  the  same  curve  referred 


to  aY'and  OY'. 

Transform   the   equation   x 
through  45°. 

Solution.    Since 

sin  45°  =  -  V2  =  — 

2  V2 


EXAMPLE 

'  —  y'^  =  Vi   when   the   axes   are   rotated 


and 


cos  45° 


1 
equations  (II)  become 


y 

7^'  y 


x'  +  ?/' 


V2  V2 

Substituting  in  the  given 
equation,  we  obtain 

or,  simplifying,  x'y'  +8  =  0, 


^; 

N 

Y' 

' 

/ 

A' 
f 

-  N 

k\ 

/ 

/ 

\ 

\ 

\ 

/ 

/ 

/ 

\ 

/ 

/ 

\ 

/ 

/ 

\ 

/ 

/ 

\ 

? 

Q? 

1 

X 

/ 

\ 

N 

/ 

\ 

\ 

/ 

/ 

\ 

\ 

/ 

/ 

l\ 

\ 

s. 

/ 

/ 

\ 

s 

\ 

PROBLEMS 

1.  Find  the  coordinates  of  the  points  (3,  1),  (—  2,  6),  and  (4,  —  1)  when 
tlie  axes  are  rotated  through  -  • 

2.  Transform  the  following  equations  when  the  axes  are  rotated  through 
tlie  indicated  angle.    Plot  both  pairs  of  axes  and  the  curve. 

Ans.  y'  —  0. 

Ans.  a;'2  =  4. 

Ans.  x'2  =  4/. 

An».  3  x'2  -  ?/2  =:  16. 

Ans.  x'2  +  2/'2  _  ^2_ 


(a)  X  -  2/  =  0  ;  -  • 

(b)  x2  +  2xj/  +  2/'-^  =  8;  -. 

^  4 

(c)  j/2  =  4x  ; 

(d)  x2  +  4x2/  +  2/2=16;  -■ 

4 
<e)  x2  +  2/2  =  r2  ;  Q. 


148 


NEW  ANALYTIC   GEOMETRY 


Ans.  V2y'"  +  4x'  =  0. 

A  ».s.  jc'-  —  4  y'-  4-  1  =  0. 

Alls.  3x"-  —  7  y"-  =  4. 

A  n.s.  1 1  x"^  +  y'-  =  22. 


If   the    axes  are 


(f)  x^  +  2xy  +  y-  +  i  X  -  i  y  =^  0  ; 

(g)  3x2  —  4x2/  —  1  :=  0  ;  arc  tan  2. 
(li)  X'  +  3xy  —  '^y-  =  2  ;  arc  tan  ] . 
( i )  x2  +  3  x(/  +  5  //-  =  1 1  ;  arc  tan  3 
( j )  3  x'^  —  3  xy  —  y-  =  5  ;  are  tan  3. 
(k)  X-  4-  4  xy  +  4  ;/■-  +  1 2  X  —  (j  y  =  0  ;  arc  tan  2. 

56.  General   transformation  of  coordinates 

moved  in  any  manner,  they 
may  l)e  brought  from  the  old 
position  to  the  new  position 
by  translating  them  to  the  new 
origin  and  then  rotating  them 
through  the  proper  angle. 

Theorem.  If  the  axes  be 
tra  us/a  ted  to  a  new  or  iff  in  (Ji,  Ic) 
and  til  en  rotated  tJirouffJi  an 
anffle  6,  tJie  f/KcifloTis  of  tlic  transfDnrKttlnn  >>/ coordinates  are 


y 

'•\ 

Y 

\ 

• 

y 

\0'(h.  kj 

.V" 

0 

y 

/^ 

y 

X 

(111) 


( X  =  :^  COS.  9  —  y^  &\n9  -Jf-  h, 


ly  =  X?  sin  6  -\-  y'  cos,  6  -\-  k. 

Proof.  To  tiunslate  the  axes  to  O'X"  and  o'Y"  we  have, 
Ijy   (I),  :r  =  .•■"  +  ]>, 

y  =  //"  +  /.•, 

where  (.r",  //")  are  the  coordinates  of  any  point  V  referred  to 
^^'A"  and  C'F". 

To  rotate  the  axes  we  set,  by  (II), 

.  ./•"  =  ,'r'  cos  Q  —  //'  sin  ^, 
//"  ;=  a-'  sin  6  +  f  cos  6. 
Substituting  these  values  of  ,'•"  and  //",  we  obtain  (HI).    Q.e.d. 

57.  Classification  of  loci.  The  loci  of  algebraic  equations 
are  classified  according  to  the  degree  of  the  ecjuations.  This 
classifiration  is  justified  by  the  following  theorem,  which 
shows  that  the  degree  of  the  ef]uation  of  a  locus  is  the  same, 
no  matter  how  the  axes  are  chosen. 


TRANSFORMATION  OF  COORDINATES  149 

Theorem.  The  degree  of  the  equation  of  a  locus  is  unchanged 
by  a  transformation  of  coordinates. 

Proof  Since  equations  (III)  are  of  the  first  degree  in  x'  and 
y',  the  degree  of  an  equation  cannot  be  raised  when  the  values 
of  X  and  y  given  by  (III)  are  substituted.  Neither  can  the 
degree  be  lowered;  for  then  the  degree  must  be  raised  if  we 
transform  back  to  the  old  axes,  and  we  have  seen  that  it  cannot 
be  raised  by  changing  the  axes.* 

As  the  degree  can  neither  be  raised  nor  lowered  by  a  trans- 
formation of  coordinates,  it  must  remain  unchanged.  o.e.d. 

58.  Simplification  of  equations  by  transformation  of  coordinates. 

The  principal  use  made  of  transformation  of  coordinates  is  to 
simplify  a  given  equation  by  choosing  suitable  new  axes.  The 
method  of  doing  this  is  illustrated  in  the  following  examples. 

EXAMPLES 
1.  Simplify  the  equation  ?/2_8x-l-62/+17  =  Oby  translating  the  axes. 
Solution.    Set  x  =  x'  -{■  h  and  y  =  y'  +  k. 
This  gives  {?/'  +  A-)^  _  8  (x'  +  /;)  +  6  (;/'  +  A:)  +  17  =  0,  or 
(1)  y'2_8x'  +  2^- 

+  0 


0. 


if  +  A;2 
-8A 
+  Qk 
+  17 

If,  now,  we  choose  for  h  and  k  such  numbers  that  the  coefficient  of  y' 
shall  be  zero,  that  is, 

(2)  2A;+6=:0, 
and  also  the  constant  term  shall  be  zero,  that  is, 

(3)  A:2-8A  +  6fc  +  17  =  0, 
the  transformed  equation  is  simply 

(4)  ?/'2_8x'  =  0. 

*  This  also  follows  from  the  fact  that  when  equations  (III)  are  solved  for 
x'  and  ?/',  the  results  are  of  the  first  degree  in  x  and  ?/. 

t  These  vertical  bars  play  the  part  of  parentheses.  Thus  2  A  +  6  is  the  coefti- 
cient  of  y'  and  A;^  -  8  /i  +  6  A:  +  17  is  the  constant  term.  Their  use  enables  us  to 
collect  like  powers  of  x'  and  7/  at  the  same  time  that  we  remove  the  parentheses 
in  the  preceding  equation. 


150 


NEW  ANALYTIC  GEOMETRY 


From  (2)  and  (3)  we  obtain  A  =  1,  fc  =  —  3,  and  these  are  the  coordinates 
of  the  new  origin. 

The  locus  may  be  readily  plotted  by  draw- 
ing the  new  axes  and  then  plotting  (4)  on 
these  axes. 

A  second  method  often  used  is  the  fol- 
lowing : 

Rewrite  the  given  equation,  collecting  the 
terms  in  ?/, 

(5)  0/  +  6  2/)  =  8x-17. 

Complete  the  square  in  the  left-hand 
member, 

(6)  (y^ +  Qy +9)  =  8x-n +  9  =  8x-8. 
Writing  this  equation  in  the  form 

(7)  {y  +  3)2  =  8{x-  1), 
it  is  obvious  by  inspection  that  if  we  substitute 
in  this  equation 

(8)  X  =  x'  +  1,     y  =  y'-  3, 
the  transformed  equation  is  y"-  =  8  x'.    But  equations  (8)  translate  the 
axes  to  the  new  origin  (1,  —  3),  as  before. 

2.  Simplify  x'^  +  4  2/'^  —  2  x  —  16 ;/  +  1  =  0  by  translating  the  axes. 
Solution.   Set  x  =  x'  ■{■  h  and  y  =  y'  +  k.    This  gives 


y 

r 

/* 

y^ 

y 

/' 

/ 

/ 

0 

/ 

X 

/ 

o 

a. 

-3; 

X' 

V 

\ 

\ 

\ 

\ 

s 

N 

(9) 


x'2  +  4y'^  +  2h 

-2 


x'+  8A; 

2/'+  h^ 

-16 

+  4fc2 

-2/i 

-16k 

+  1      1 

=  0. 


Let  us  choose  the  new  origin  so  that  in  (9)  the  coefficients  of  x'  and  y' 
shall  be  zero  ;  that  is,  so  that 

(10)  2  /i  -  2  =  0   and   8  fc  -  16  =  0. 

From  (10),  h  =  1,  k  —  2,  and  these  values  substituted  in  (9)  give  the 
transformed  equation 

(11)  x'2 -I- 4  ?/'2  =  16. 

The  locus  of  the  given  equation  is  now  readily  drawn  by  constructing 
parallel  axes  through  (1,  2)  and  plotting  equation  (11)  on  these  axes. 
A  second  method  is  the  following : 
Collect  corresponding  terms  in  the  given  equation  thus  : 

(12)  (x2-2x)  +  4(y2_42/)  =-1. 


TRANSFORMATION  OF  COORDINATES 


151 


Complete  the  squares  within  the  parentheses,  adding  the  corresponding 
numbers  to  the  right-hand  member, 

(13)  (a;2-2x  +  l)  +  4(2/2-42/  +  4) 

=  -1  +  1  +  16  =  16. 
Writing  (13)  in  the  form 

(a:_l)2  4.4(^_2)2  =  16, 
it    is    obvious    by    inspection    that    by 
substituting 

(14)  x  =  x'+l,     j/  =  y'  +  2, 
the  simple  new  equation  x"^  -if  Ay"^  =  16  results.    But  equations  (14)  trans- 
late the  axes  to  the  new  origin  (1,  2),  the  same  as  in  the  first  method. 

3.  Kemove  the  x?/-term  from  x^  +  4  jy  +  ?/-  =  4  by  rotating  the  axes. 
Solution.    Set  x  =  x'  cos  0  —  y'  sin  6  and  y  =  x'  sin  ^  +  if  cos  0,  whence 


r|r| 

^ 

r 

'—"I 

■^ 

/ 

\ 

V 

0' 

ri, 

I) 

/ 

x' 

>^ 

U^ 

0 

X 

+  4  sin  6  cos  6 
+  sin'-  d 


x"-  —  2  sin  9  cos  6 
+  4(cos2^-sin26') 
+  2  sin  0  cos  9 


x'y'  +  sin2  0 

—  4  sin  9  cos  9 
+  cos2  9 


V 


4, 


or,  since  2  sin  9  cos  ^  =  sin  2  ^  and  cos^  9  —  sin2  9  =  cos  2  9^ 

(15)         (1  +  2  sin  2  ^)  x'2  +  4  cos2  (9  •  xV  +  (1  -  2  sin 2  (9)2/'2  =  4. 

The  new  equation  is  to  contain  no  x'y'-term.     Hence,  setting  the 

coefficient  of  x'y'  equal  to 

zero, 

cos  2^  =  0. 

.-.  2^  =  -     and     ^  =  -. 
2  4 

Substituting     in      (15), 

since  sin  —  =  1,  the  trans- 

2 
formed  equation  is 

3x'2_y'2^4. 

The  locus  of  this  equation 
is  the  hyperbola  plotted  on 
the  new  axes  in  the  figure. 

These    examples    show 
that  it  is  often  wise  not  to 
plot  the  locus  of  an  equation  as  it  stands,  but  ratlier  to  endeavor  first 
to  simplify  by  transformation  to  new  axes. 


.N 

. 

1" 

/x 

^ 

\ 

\ 

/ 

\ 

\ 

/ 

\ 

\ 

/ 

\ 

V 

/ 

\ 

\ 

/ 

~ 

\ 

/ 

\ti 

^ 

\ 

\ 

u 

Si 

Y. 

X 

/ 

\ 

\ 

^ 

. 

/ 

\ 

/ 

k 

/ 

\ 

\ 

/ 

\ 

\ 

/ 

\ 

\ 

/ 

\ 

k 

Ans. 

X''  +  4  2/' 

=  0. 

Ans. 

x'2  =  4  y'. 

Ans. 

X'2  +  2/'2  : 

=  16. 

Ans. 

y'2  =  6x' 

Ans. 

X'2  -  2/'2  : 

=  2. 

Ans. 

X'2  +  4  ij"- 

2  =  16, 

Ans. 

8x'+?/3 

=  0. 

152  NEW  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Simplify  tlie  following  equations  by  translating  the  axes.  Plot  both 
pairs  of  axes  and  the  curve. 

(a)  x2  +  6x  +  4?/  +  8  =  0. 

(b)  x2-42/  +  8  =  0. 

(c)  x2  +  2/2  +  4x  -  6 y  -  3  =  0. 

(d)  2/2  _  6x  -  lOy  +  19  =  0. 

(e)  x2  -  2/2  +  8x  -  14  ?/  -  35  =  0. 

(f)  x2  +  42/2 -16x  +  24  2/  +  84  =  0. 

(g)  2/3  +  8x-  40  =  0. 

2.  Remove  the  xy-term  from  the  following  equations  by  rotating  the 
axes.    Plot  both  pairs  of  axes  and  the  curve. 

(a)  x2  —  2x2/ +  2/^  =  12.  Ans.  %j'^  =  Q. 

(b)  x2  —  2  X2/  +  2/^  +  8  X  +  8  2/  =  0.  Ans.  V2  y'"  +  8  x'  =  0. 

(c)  xy  =  18.  Ans.  x'2  -  ?/2  =  36. 

(d)  25  x2  +  14  X2/  +  25  ?/  =  288.  Ans.  16  x'2  +  9  2/'2  =  144. 

(e)  3  x2  —  10x2/  +  3  2/2  =  0.  ^ns.  x'2  _  4  2/'2  =  0. 

3.  Translate  the  axes  so  that  each  of  the  following  equations  is  trans- 
formed into  a  new  equation  without  any  terms  of  the  first  degree  in  the 
new  coordinates.   Draw  the  locus. 

(a)  x2  —  4  X2/  +  6  2/  =  0.      Ans.  /i  =  f ,  k  —  \. 

(b)  2/2- 2x2/ +  3x  =  0.  (e)  3x2  —  x?/ —  2/2  +  4x  =  0. 

(c)  x2  +  X2/  +  2/2  +  6x  =  0.  (f)  2x2/  +  6x-8  2/  =  0. 

(d)  x2  -  x?/  +  2 2/2  +  62/  =  0.  (g)  3xj/  +  4 2/  -  2  =  0. 


CHAPTER  X 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 

59.  The  parabola.    Consider  the  following  locus  problem. 

A  point  moves  so  that  its  distances  from  a  fixed  line  and  a 
fixed  point  are  equal.    Determine  the  nature  of  the  locus. 

Solution.  Let  DD'  be  the  fixed 
line  and  F  the  fixed  point.  Draw 
the  a;-axis  through  F  perpendicular 
to  DD'.  Take  the  origin  midway 
between  F  and  DD'. 

Let 

(1)  distance  from  F  to  DD' =  p. 

Then,  if  P  (x,  y)  is  any  point  on 
the  locus, 

(2)  FP  =  MP. 

But      FP  =  V(.T  —  ^py-hV%     MP  =  MN  -{-  Nl'  =  ^p  +x. 
Substituting  in  (2), 

^{x-\py  +  if=\p  +  x. 

Squaring  and  reducing, 

(3)  if  =  2px. 

The  locus  is  called  a, parabola.  The  fixed  line  DD'  is  called  the 
directrix,  the  fixed  point  F,  the  focus.  From  (3),  it  is  clear  that 
the  .X-axis  is  an  axis  of  symmetry.  For  this  reason,  the  x-axis 
is  called  the  axis  of  the  parabola.  Furthermore,  the  origin  is  on 
the  curve.  This  point,  midway  between  focus  and  directrix,  is 
called  the  vertex. 

1.58 


1) 

Y. 

N 

^^ 

M 

y 

D 

0 

Y 

V   ^(^,0 

)                   X 
FP=MP 

154 


NEW  ANALYTIC  GEOMETRY 


Theorem.    If  the  origin  is  the  vertex  and  the  x-axis  the  axis 
of  a  parabola,  then  its  equation  is 


(I) 


y^  =  2px. 

and  the  equation  of  the  directrix 


^,0 


The  focus  is  the  point 

P 

IS  X  ^ 

2 
A  discussion  of  (I)  gives  us  the  following  properties  of  the 
parabola  in  addition  to  those  already  obtained. 

1.  Values  of  x  having  the  sign  opposite  to  that  of  2^  are 
to  be  excluded.  Hence  the  curve  lies  to  the  right  of  YY^ 
when  p  is  positive  and  to  the  left  when  ^;  is  negative. 

2.  No  values  of  y  are  to  be  excluded ; 
hence  the  curve  extends  indefinitely  up 
and  down. 

The  chord  drawn  through  the  focus 
parallel  to  the  directrix  is  called  the 
latus  rectum.  To  find  its  length,  put 
X  =  \p  in  (I).  Then  ?/  =±p,  and  the 
length  of  the  latus  rectum  =  2^? ;  that 
is,  equals  tlie  coefficient  ofx  in  (I). 

It  will  be  noted  that  equation  (I)  contains  two  terms  only ; 
namely,  the  sqicare  of  one  coordinate  and  tJoe  first  power  of  the 
other.    Obviously,  the  locus  of 

x^  =  2py 
is  also  a  parabola,  and  thus  we  have  the 

Theorem.    If  tlie  origin  is  the  vertex  and  the  y-axis  the  axis 
of  a  parabola^  then  its  equation  is 
(II)  x'  =  2py. 

The  fociis  is  the  point  [  0,  -  )>  and  the  equation  of  the  directrix 

is  y  =— -• 
-^  2 

Equations  (I)  and  (II)  are  called  the  typical  fo)'ms  of  the 

equation  of  the  parabola. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


155 


Equations  of  the  forms 

Ax^  +  Ei/  =  0   and    Cif  —  Dx  =  Q, 

where  A,  E,  C,  and  D  are  different 
from  zero,  may,  by  ti'ansposition  and 
division,  be  written  in  one  of  the 
forms  (I)  or  (II), 

To  plot  a  parabola  quickly  from 
its  typical  equation,  its  position  (above 
or  below  A' A'',  to  the  right  or  left  of 
YY')  is  best  determined  by  discussion 

of  the  equation.    The  value  of  2p  is  found  by  comparison 
with  (I)  or  (II),  and  the  focus  and  directrix  are  then  plotted. 


EXAMPLES 

1.  Plot  the  locus  of  x^  +  4y  =  0  and  plot  the  focus  and  directrix. 

Solution.   The  given  equation  may  be  written 

x^  =  —  4  y. 

The  t/-axis  is  an  axis  of  symmetry  ;  positive  values  of  y  must  be  ex- 
cluded. Hence  the  parabola  lies  below  the  a;-axis.  The  table  gives  a  few 
points  on  the  curve. 


X 

y 

0 

±2 
±4 

0 

- 1 

-4 

5M 

> 

D 

,-[ 

D 

0 

X' 

^ 

N 

X 

/ 

-'•(O.-ll 

\ 

/ 

\ 

/ 

\ 

t 

' 

I 

1 

\ 

y 

Comparing  with  (II),  p  =  —  2.  The  focus  is  therefore  the  point 
(0,-1)  and  the  directrix  the  line  ?/  =  l.  The  length  of  the  latus 
rectum  is  4.  Every  point  on  the  locus  is  equidistant  from  (0,  —  1)  and 
the  line  y  =  '[. 


156 


NEW  ANALYTIC   GEOMETRY 


2.  Find  the  equation  of  the  parabola  whose  focus  is  (4,  —  2)   and 
■directrix  the  line  x  =  l. 

Solution.  In  the  figure,  by  definition, 
.(1)  FP  =  FM. 

^  But  FP 

and 


V{x  -  if  +  {y  +  2)2, 
PJlf  =  x-l. 
Substituting  in  (1)  and  reducing, 

(2)  2/2  — 6x  +  42/  +  19  =  0.    Ans. 

If  tlie  axes  are  translated  to  the 
vertex  (|,  —  2)  as  a  new  origin,  that  is, 
if  we  substitute  in  (2)  x  =  x'  +  |  and 
y  =  y'  —  2,  the  equation  reduces  to 
the  typical  form  ?/'2  —  6  x'  =  0. 

A  second  and  useful  method  is  the 
following : 

Draw  the  axis  VX'  of  the  parabola 
and  the  tangent  VY'  at  the  vertex. 
Referred  to  these  lines  as  temporary 
axes,  the  equation  must  have  the 
typical  form 

(3)  2/2  =  6  X, 
since  p  =  3. 

Now  translate  the  temporary  axes  so 
that  they  will  coincide  with  the  given 
axes.  The  coordinates  of  0  referred  to 
the  temporary  axes  are  (  —  f ,  2).  Sub- 
stituting in  (3)  x=x'— I,  y  =  y'-\-  2,  and 
reducing,  we  obtain  the  equation  (2). 


PROBLEMS 

1.  Plot  the  locus  of  the  following  equations.     Draw  the  focus  and 
directrix  in  each  case  and  find  the  length  of  the  latus  rectum. 

(a)  2/2  =  4x.  (d)  y^-Qx-  0. 

(b)2/2  4-4x  =  0.  (e)  x2 +  102/  =  0. 

(c)x2-82/  =  0.  (f)2/^  +  x  =  0. 

2.  Find  the  equations  of  the  following  parabolas  : 

(a)  directrix  x  =  0,  vertex  (3,  4).  Ans.    (y  —  4)2  =  12  (x  —  3). 

(b)  focus  (0,  -  3),  vertex  (2,  -  3).  Ans.   y^  +  8x  +  6y -7  =  0. 


PARABOLA,  ELLIPSE,  AND  HrPERBOLA 


157 


(c)  axis  X  =  0,  vertex  (0,  —  4),  passes  through  (6,  0).  Ans.  x"  =  9y  +  86. 

(d)  axis  y  =  0,  vertex  (6,  0),  passes  through  (0,  4). 

Ans.  3y"  =  48-  8x. 

(e)  directrix  x  +  2  y  —  1  =  0,  focus  (0,  0). 

Ans.  (2  X  —  ?/)2  +  2  X  +  4  ?/  -  1  =  0. 

3.  Transform  each  of  tlie  following  equations  to  one  of  the  typical 
forms  (I)  or  (II)  by  translation  of  the  axes.    Draw  the  figure  in  each  case. 


(a)  ?/2  +  4x  +  4y-2  =  0. 

(b)  x2+  6x  +  ?/-2  =  0. 

(c)  x2+3x  +  4?/-l=0. 

(d)  y2+5x  +  8y  =  0. 

(e)  2x-+  5y  +  4  =  0. 

(f)  2/2+  6x- 9  =  0. 

(g)  7x2 +  8y  + 10  =  0. 
(h)  x2  +  4  y/  +  4  =  0. 


Ans.    y'"  +  4x'  =  0. 
Ans.   x'2  -f-  2/'  =  0. 


(i)  2?/  +  3x  —  8  =  0. 
(j)  5x2  +  102/ +  12  =  0. 
(k)  3x2-6?/  +  8  =  0. 
(1)  2x2- Ox  +  2/  =  0. 


4.  Show  that  abscissas  of  points  on  the  parabola  (I)  are  proportional 
to  the  squares  of  the  ordinates. 

5.  Find  the  equation  in  polar  coordinates  of  a  parabola  if  the  focus  is 
the  pole,  and  if  the  axis  of  the  parabola  is  the  polar  axis. 

A  P 

Ans.    p= i 

l-cos^l 

60.   Construction  of  the  parabola.    A  parabola  Avhose  focus  and 
directrix  are   given  is    readily 
constructed  by  rule  and  com- 
passes as  follows : 

Draw  the  axis  MX.  Con- 
struct the  vertex  V,  the  middle 
point  of  MF.  Through  any 
point  A  to  the  right  of  V  draw 
a  line  AB  parallel  to  the  direc- 
trix. From  F  as  a  center  with 
a  radius  equal  to  MA  strike 
arcs  to  intersect  AB  at  P  and 
Q.  Then  P  and  Q  are  points 
on  the  parabola.  For  FP  =  MA,  by  construction,  and  hence. 
P  is  equidistant  from  focus  and  diivctrix. 


158 


NEW  ANALYTIC   GEOMETRY 


C     a'   I) 


n 

VI 

I 

1 

1 

1 

1      '    ^^^^ 
1      [      1  ""' 

1      1      1 
1      1      ' 

1 

r^^^ 

t 

1 

A 

a 

\}_ 

c 

-^-^pan 

— > 

B 

a,  h,  c, 

I,  m,  n. 

By  changing  the  position  of  A  we  may  construct  as  many 
points  on  the  curve  as  desired. 

Parabolic  arch.  When  the  span  AB  and  height  OH  of  a  para- 
bolic arch  are  given,  points  on  the  arch  may  be  constructed  as 
follows  : 

Draw  the  rectangle  A  BCD. 

Divide  ^//and  AC  into  the 
same  number  of  equal  parts. 

Starting  from  A,  let  the  suc- 
cessive points  of  division  be 

on  AH, 

on  A  C, 

Now  draw  the  perpendicular  aa'  to  AB,  and  draw  01.  Mark 
the  intersection.  Do  likewise  for  the  points  b  and  vi,  c  and  n. 
The  intersections  are  points 
on  the  parabola  required. 

Proof.    Take   axes    OX 
and  OY,  as  in  the  figure. 
Let 
(1)    OM'  =  X,    M'P  =  ij, 

AB  =  2a,   OH  =  h.  Y 

By  construction,  NC  and  MH  are  equal  parts  of  ^C  and  AH 

respectively. 

.^  .    NV_MH  NC  _x 

^"^  '  ■  'aC  ~  Ih'     ^^       h    ~  a 

From  the  similar  triangles  OM'P  and  OCN, 
y  _NC  _NC 


c 

il/'p a^- 

0 

N 

\\ 

X 

B(c 

A(r 

a,h)M            H 

\ 

I' — 

— a ^> 

(3) 


OC 


Substituting  the  value  of  NC  from  (2)  into  (3),  and  reducing, 


(4) 


2  * 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


159 


This  is  the  typical  form  (II),  and  the  locus  passes  through 

0,  A  (—  a,  h)  and  B(a,  Ji),  as  required. 
Solving  (4)  for  y,  we  get 


(5) 


^       a}     ' 


X 

la 

ia 

fa 

a 

y 

-h  h 

ih 

T%/' 

h 

showing  that  y  varies  as  the 

square  of  x,  and  giving  a  simple  formula  for  computing  y,  as 

in  the  table. 

61.  Equations  (I)  and  (II)  are  extraordinarily  simple  types 
of  equations  of  the  second  degree.    The  question, 

To  derive  a  test  for  determining  if  the  locus  of  a  given  equa- 
tion of  the  second  degree  is  a  parabola, 

will  be  answered  in  Art.  70 ;  but  at  this  point,  if  the  previous 
results  are  borne  in  mind,  we  may  state  the 

Theorem.  The  locus  of  an  equation  of  the  second  degree  is  a 
parabola  if  the  only  tervi  of  the  second  degree  *  is  the  square  of 
one  coordinate,  and  if  also  the  first  power  of  the  other  coordi- 
nate is  present  in  the  equation. 

For  illustration,  see  Problem  3,  p.  157. 

62.  The  ellipse.    Let  us  solve  the  following  locus  problem : 
Given    two    fixed   points   F  and 

F'.  A  point  P  moves  so  that  the 
sum  of  its  distances  from  F  and 
F'  remains  constant.  Determine  the 
nature  of  the  locus. 

Solution.  Draw  the  ar-axis  through 
F  and  /•",  and  take  for  origin  the 
middle  point  of  F'F.    By  definition, 

(1)  PF  +  PF'  =  a  constant. 


PF+  PF'=2a 


*The  student  should  not  forget  tiiat  the  product  zy  is  a  term  of  the  second 
degree. 


160 


NEW  ANALYTIC   GEOMETRY 


Let  us  denote  this  constant  by  2  a.    Then  (1)  becomes 
(2)  PF  +  PF'  =  2  a. 

Let  FF'  =  2c.    Then 


PF  =  ■\/{x  —  (■)'-  +  2/^     PF'  =  V(cc  +  cf  +  y*, 
since  the  coordinates  of  F  are  (c,  0),  and  of  F',  (—  c,  0). 
Hence  (2)  becomes 

(3)  V(x-  -  c)^  +  f  +  V(at  +  cy  +  if  =  2  a. 
Transposing  one  of  the  radicals,  squaring  and  reducing,  the 

result  is 

(4)  {a'  -  c^  X-  +  ahf  =  d\a?  -  <P). 
For  added  simplicity,  set  * 

(5)  a'-c'  =  h\ 
Then  (4)  becomes  the  simple  equation 

(6)  Jj'^x^  +  a^ij-  =  a%-. 
Discussion.    The  intercepts  are, 

on  XX',  ±a;  on  YY',  ±  h. 
The  axes  XX'  and  YY'  are  axes  of  symmetry  and  O  is  a  center 
of  symmetry. 

Solving  (6)  for  x  and  for  y, 

x  =  ±^^-y/V'  -if, 


V 


=  ±^V.7 


Hence  the  values  of  x  can- 
not exceed  a  numerically, 
nor  can  the  values  of  y 
exceed  b  numerically.  The  curve  is  therefore  closed. 

The  locus  is  called  an  ellipse.  The  point  0,  which  bisects 
every  chord  passing  through  it,  is  called  the  center.  The  given 
fixed  points  F  and  F'  are  called  the  foci.    The  longest  chord 

*  This  is  permissible.   For  PF+  PF'  >  FF',  or  2  a  >  2  c ;  that  is,  a  >  c,  and 

q2  _  (.2  jg  2i  positive  number. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


161 


AA '  through  O  is  called  the  major  axis  ;  the  shortest  chord  BB', 

the  minor  axis.    Obviously, 

(7)  major  axis  =  2  a,     minor  axis  =  2h.  , 

Dividing  (6)  through  by  a^"^,  and  summarizing,  gives  the 

Theorem.    The  equation  of  an  ellipse  whose  center  is  the  orirjin 
and  whose  foci  are  on  the  x-axis  is 


(III) 


where  2  a  is  the  major  axis  and  2  b 
the  minor  axis.  Ifc^  ^a^  —  6^,  then 
the  foci  are  (±  c,  0). 

If  the  foci  are  on  the  ?/-axis, 
and  if  we  keep  the  above  nota- 
tion, the  equation  of  the  ellipse  is 
obviously 

(8)  a^x^  +  by  =  a%%     or 

Equations  (6),  (8),  and  (III)  are  typical  equations  of  the 
ellipse,  and  are  of  the  form 

(9)  Ax"^  -\-Bf=C, 
where  ^4,  2?,  and  C  agree  in  sign. 

In  the  figure  'BF'^  =  h'^  +  c\ 
Substituting  the  value  of  c^  from 
(5),  then  BF  =  a\  Hence  the 
property  :  The  distance  from  either 
focus  to  the  end  of  the  minor  axis 
equals  the  sem  imajor  axis. 

The  chord  drawn  through  either  focus  perpendicular  to  the 
major  axis  is  called  the  latus  rectum.  Its  length  is  deter- 
mined by  setting  a;  =  c  in  (III),  and  solving  for  y.    This  gives 

1/  =  -  Va**  —  c^  =  —  •    Hence 


F; 

, 

^^ 

B 

^-^c,#) 

> 

A 

h 

\  i\ 

/ 

C  \[       \ 

^'\ 

o 

\F  JU-^ 

X 


(10) 


length  of  latus  rectum  = • 


162 


NEW  ANALYTIC  GEOMETRY 


Eccentricity.  When  the  foci  are  very  near  together  the  ellipse 
differs  but  little  from  a  circle.  The  value  of  the  ratio  OF :  OA 
may,  in  fact,  be  said  to  determine  the  divergence  of  the  ellipse 
from  a  circle.  The  value  of  this  ratio  is  called  the  eccentricity 
of  the  ellipse,  and  is  denoted  by  e.    Hence 

OF       c 

The  value  of  e  varies  from  0  to  1.  If  the  major  axis  A  A' 
remains  of  fixed  length,  then  the  "  flatness  "  of  the  ellipse  in- 
creases as  e  increases  from  0  to  1,  the  limiting  forms  being  a 
circle  of  diameter  ^A'  and  the  line  segment  ^1.4'. 

From  (11)  and  (5), 


(12) 


b^  =  a'-c'  =  a\l-e^). 


To  draw  an  ellipse  quickly  when  its  equation  is  in  the  typical 
form,  proceed  thus : 

1.  Find  the  intercepts,  mark  them  off  on  the  coordinate 
axes,  and  set  the  larger  one  equal  to  a,  the  smaller  equal  to  h. 
Letter  the  major  axis  A  A'  and  the  minor  axis  BB'. 

2.  Find  c  from  c^  =  a^  —  V^.  Mark  the  foci  F  and  F'  on  the 
major  axis. 

3.  Calculate  directly  one  or  more  sets  of  values  of  the  coordi- 
nates, and  sketch  in  the  curve. 


EXAMPLE 

Draw  the  ellipse  ix^  +  2/^  =  16. 

Solution.  The  intercepts  are,  on  XX',  ±  2 ;  on 
YY',  ±4.  Hence  the  major  axis  fallson  YY',  and 
a  =  4,  6  =  2,  c  =  V12  =  2  VS  =  3.4.  The  foci  are 
on  the  2/-axis.    The  length  of  the  latus  rectum 

equals =  2.  The  eccentricity  e  —  ~  =\  v 3.    ^ 

The  points  found  in  the  table 
ai-e  the  ends  of  the  latus  rectum. 
If  P  is  any  point  on  the  ellipse, 
then  PF  +  PF'  =  2  a  =  8. 


X 

y 

±1 

±3.4 

PARABOLA,  ELLIPSE,  AN^D  HYPERBOLA  163 

PROBLEMS 

*  1.  Plot  each  of  the  following  equations.  Letter  the  axes  and  mark 
the  foci.  Find  the  eccentricity,  the  length  of  the  latus  rectum,  and 
draw  the  latus  rectum. 

' (a)  a;2  +  92/2  =:  9.  -(e)  9x^  +  iy^  =  36. 

' (b)  9x2  +  16 2/2  =  144^  (f  j  2 x2  +  2/2  =  25. 

(c)  2 x2  +  2/2  =  4.  (g)  4 x2  +  8 2/2  =  32. 

(d)  4x2  +  Qy2  ^  36.  (h)  7^2  +  3j/2^  21. 

•  2.  Transform  each  of  the  following  equations  by  translation  of  the 
axes  so  that  the  transformed  equation  shall  lack  terms  of  the  first  degree 
in  the  new  coordinates.    Draw  the  figure. 

.  (a)  x2  +  4  2/2  +  6  X  —  8  2/  =  0.  Ans.   x'"-  +  4  y'-  =  13. 

.(b)  9x2  +  4y2  +  30j.  _  4y  +  1=0. 

(c)  x2  +  5  2/2  +  102/  =  20. 

(d)  5x2  +  2/2  +  lOx  +  4  2/  =  6. 

(e)  3  x2  +  2/2  +  6  X  —  4  ?/  =  2. 

.  (f)  4x2  +  52/2  +  4x  +  20  2/  =  20. 

3.  Find  the  equation  of  each  of  the  following  ellipses: 

(a)  major  axis  =  8,  foci  (5,  2)  and  (—  1,  2). 

Ans.    7  (x- 2)2  + 16  (2/ -2)2  =  112. 

(b)  major  axis  =  10,  foci  (0,  0)  and  (0,  6). 

Ans.   25x2  +  16(2/- 3)2  =  400. 

(c)  minor  axis  =  8,  foci  (—1,0)  and  (4,  0). 

(d)  minor  axis  =  4,  foci  (0,  —  2)  and  (0,  4). 

63.  Construction  of  the  ellipse.    The  definition  (2)  of  the  pre- 
ceding section  affords  a  simple  method  of  drawing  an  ellipse. 

Place  two  tacks  in  the  drawing  board  at 
the  foci  F  and  F'  and  wind  a  string  about 
them  as  indicated.  If  now  a  pencil  be  placed 
in  the  loop  FPF'  and  be  moved  so  as  to 
keep  the  string  taut,  then  J^F  -f  PF'  is 
constant  and  P  describes  an  ellipse.  If  the 
major  axis  is  to  be  2  a,  then  the  length  of  the  loop  FPF'  must 
be  2  a  +  2  c. 


164 


NEW  ANALYTIC   GEOMETRY 


A  useful  construction  of  an  ellipse  by  rule  and  compasses  is 
the  following: 

Draw  circles  on  the  axes  A  A'  and  BB'  as  diameters.  From 
the  center  0  draw  any  radius  intersecting  these  circles  in  M 
and  N  respectively.  From  M  draw 
a  line  MB  parallel  to  the  minor 
axis,  and  from  N  a  line  JVS  paral- 
lel to  the  major  axis.  These  lines 
will  intersect  in  a  point  P  on  the 
ellipse. 

Proof.      Take    the     coordinate 
axes  as  in  the  fiq-ure  below.    Let 


OA  =  x,     AP  =  y  =  OD. 
Clearly,  OB 

OC 


ZMOX  =  cf>. 


semimajor  axis  =  a, 
semiminor  axis  =  b. 


Then  in  the  right  triangle  OAB, 


(1) 


OA       X 


Similarly,  in  the  right  triangle  ODC,  AOCD  =  ZCOA  =  ^, 
and 

(2)  «in*  =  ^^f■ 

But  cos^  <f>  +  siu^-  ^  =  1.   Hence, 

2  2 

from  (1)  and  (2),  -^  +  '^  =  1,  and 

P(x,  y)  lies  on  the  ellipse  whose 
semiaxes  are  a  and  b.  Q.e.d. 

The  angle  ^  is  called  the  eccen- 
tric angle  of  P. 

The   construction    circles    used 
in  this  problem  are  called,  respectively,  the  major  and  minor 
auxiliary  circles. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  165 

64.  Equations  (6)  and  (8)  of  Art.  62  are  simple  equations  of 

the  second  degree.    We  may  ask  the  question, 

What  is  the  test  that  the  locus  of  a  given  equation  of  the 
second  degree  shall  be  an  ellipse  ? 

Reserving  for  a  later  section  the  answer  to  this  question,  we 
have,  however,  some  light  on  it  now.  For  we  have  observed 
in  Problem  2,  p.  163,  that  the  locus  was  in  each  case  an  ellipse. 
These  equations  agree  in  the  respect  that  there  is  no  xy-term, 
and  the  squares  of  x  and  y  have  unequal  positive  coefficients. 
Consider  such  an  equation,  for  example, 

(1)  x'+^y'+^x-^y +  N=0, 

where  N  is  some  nun^ber.    If  we  translate  the  axes  to  the  new 
origin  (—  2, 1),  the  transformed  equation  is 

(2)  a;''+4?/'^=8-iV. 

If  N  is  less  than  8,  the  locus  is  an  ellipse. 
If  N  =  8,  the  locus  is  the  single  point  (0,  0),  often  called  a 
point-ellipse. 

If  N  is  greater  than  8,  there  is  no  locus. 

This  discussion  is  general,  and  may  be  summarized  in  the 

Theorem.  If  an  equation  of  the  second  degree  contains  no 
xy-term,  oMd  if  x?  and  y^  occur  with  coefficients  having  like 
signs,  the  locus  is  necessarily  an  elVqyse  or  point-ellipse. 

The  case  when  x^  and  y^  have  equal  coefficients  has  been  dis- 
cussed in  Art.  38.  The  circle  and  point-circle  may,  of  course, 
be  regarded  as  special  cases  of  the  ellipse  and  point-ellipse. 

65.  The  hyperbola.  Let  us  next  turn  our  attention  to  a  third 
locus  problem. 

Given  two  fixed  points  F  and  F'.  A  point  7'  moves  so  that 
the  difference  of  its  distance  from  F  and  F'  remains  constant. 
Determine  the  nature  of  the  locus. 


166 


NEW  ANALYTIC   GEOMETRY 


Solution.    Draw   the  x-axis  through  the  fixed   points,  and 
take  for  origin  tlie  middle  point  of  F'F.    By  definition 

(1)  PF'  —  PF  =  a  constant. 

Let  us  denote  tliis  constant 
by  2  a.    Then  (1)  becomes 

(2)  PF'  -  PF=2a. 
Let        FF'  =2c. 


Tlien    PF  =  ^{x  —  cf-{-if, 
and  PF  =  V(^  +  cf  +  if, 
since  the  coordinates  of  F  are 
(c,  0),  and  of  F' ,  (-  c,  0). 

Substituting  in  (2), 

(3)  V(.r  +  if  +  if  -  V(a;  -  cf  +  f  =  2  a. 
Transposing  either  radical,  squaring  and  reducing,  the  result  is 

(4)  (a^  —  c^)  X-  +  a^y'  =  a^  (a^  —  c^). 

For  added  simplicit}',*  set 

(5)  a^  —  c^  =  —  Jy-,     or     <?  —  a^  =  IP: 
Then  (4)  becomes  the  simple  equation 

(6)  Irx^  -  aSf  =  d'U\ 

Discussion.  The  intercepts 
are,  on  XX',  ±  a;  on  YY', 
±  h  V-^ ;  that  is,  the  locus  does 
not  cross  the  y-axis.  The  coef- 
ficient of  the  V— 1  in  the  im- 
aginary intercept  on  the  y-axis 
is,  however,  h.  The  axes  XX' 
and  YY'  are  axes  of  symmetry 
and  O  is  a  center  of  sjnnmetry. 


*This  is  permissible.  For  in  the  figure,  FF'-  PF<F'F,  or  2a  <2c  ;  that 
is,  a<c,  and  a^  -  c^  is  a  negative  number. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


167 


Solving  (6)  for  x  and  for  3/, 


whence  we  conclude  that  all  values  of  x  between  —  a  and  a 
must  be  excluded,  but  no  values  of  y. 

When  X  increases,  y  also  increases,  and  the  curve  extends 
out  to  infinity,  consisting  of  two  distinct  branches.* 

The  locus  is  called  a  hyperbola,  the  point  0,  which  bisects 
every  chord  drawn  through  it,  is  called  the  center.  The  given 
fixed  points  F  and  F'  are  the  foci.  The  chord  A  A'  is  named 
the  transverse  axis.  Marking  off  on  YY'  from  0  the  lengths  ±  b, 
the  line  BB'  (Fig.  p.  166)  is  called  the  conjugate  axis.    Thus  the 

(7)  transverse  axis  =  2  a,    conjugate  axis  =  2  b. 
Dividing  (6)  through  by  a%^,  and  summarizing,  gives  the 
Theorem.    The  equation  of  a  hyperbola  v:hose  center  is  the 

origin  and  whose  foci  are  on  the 

x~axts  ts       yH      j^2 

(IV) 

whert  2  a' is  the  tra.nsverse  axis 
and  2  b  the  conjugate  axis.  If 
&  ■=  c^  -\-  6^,  then  the  foci  are 
(±  c,  0). 

If  the  foci  are  on  the  ?/-axis, 
and  if  we  preserve  the  notation, 
the  equation  of  the  hyperbola  is 
obviously 

(8)  aV  -  iV  =-  a%\     or     ^  - 

Equations  (6)  and  (8)  are  typical  equations  of  the  hyper- 
bola.   They  are  of  the  f oi  m 

(9)  Ax'^Bf^C, 
where  A  and,  1',  differ  in  sign. 


7^      6^~    ' 


*On  the  left-hand  branch,  (2)  is  replaced  by  PF~PF'=2a. 


168 


NEW  ANALYTIC   GEOMETRY 


In  the  preceding  figures  AB  =a^  -\-h^.  Substituting  the 
value  of  IP'  from  (5),  AB  =  c^.  Hence  the  property :  The 
distance  between  the  extremities  of  the  axes  equals  half  the 
distance  between  the  foci. 

The  chord  drawn  through  a 

focus  and  perpendicular  to  the 

transverse  axis  is  called  the  latus 

rectum.    We  may  determine  its 

length  by  setting  a;  =  c  in  (IV) 

and  solving  for  y.  Thus,  by  (5)  we 

,      .  b    r^-, 5  h' 

obtain  2/  =  ±  -Vt-  —  «-  =  ±  "• 

Hence 


(10)  length  of  latus  rectum  = 


2&2 


Eccentricity.  The  value  of  the  ratio  OF:  OA  in  the  hyperbola 
is  called  the  eccentricity/  of  the  curve,  as  in  the  case  of  the 
ellipse.    Denoting  the  eccentricity  by  e,  then 

OF       c 


For  a  hyperbola,  e  >  1.  The  relation  of  the  value  of  e  to  the 
shape  of  the  curve  will  be  made  clear  later.    From  (5)  and  (11), 

(12)  b'^  =  c"  -a^  =  cr  {tP  - 1). 

To  draw  a  hyperbola  quickly  when  its  equation  is  in  the 
typical  form  (9),  proceed  thus  : 

1.  Find  the  intercepts  and  mark  them  off  on  the  proper  axis. 
Set  a  equal  to  the  real  intercept  and  b  equal  to  the  coefficient 
of  V— 1  in  the  imaginary  intercept.  Lay  off  the  conjugate  axis  ; 
letter  it  BB'  and  the  transverse  axis  A  A '. 

2.  Find  c  from  c^  =  a^  +  b\  Mark  the  foci  F  and  F*  en  the 
transverse  axis. 

3.  Calculate  directly  one  or  more  sets  of  values  of  the  coor- 
dinates, and  sketch  the  curve. 


PARABOLA,  ELLIPSP:,  AND  HYPERBOLA 


169 


EXAMPLE 


Draw  the  hyperbola 

4x2-  52/2  +  20  =  0. 


Solution.    The  intercepts  are,  on 

XX\  ±  V  -  5  =  ±  VS  V^  ;  on 
YY\  ±2.  Hence  6  =  Vs,  a  =  2, 
c  =  V a2  -|-  62  =  3,  and  the  transverse 
axis  and  the  foci  are  on  YY' .  The  ec- 
centricity is  f .  The  length  of  the  latus 
2  62 


rectum  is 


5. 


If  P  is  any  point  on 
the  hyperbola,  then 
-p-E"  -  PF  =  'i. 


X 

y 

0 

±2 
±3 

PROBLEMS 

1.  Plot  each  of  the  following  equations,  letter  the  axes,  and  mark  the 
foci.  Find  the  eccentricity,  the  length  of  the  latus  rectum,  and  draw  the 
latus  rectum. 


(a)  5x2-4  2/2  =  20. 

(b)  x2  -  8  2/2  +  8  =  0. 

(c)  9  x2  -  2/2  _  9. 

(d)  3x2-2/2  =  12. 


(e)  x2  -32/2  +  3  =  0. 

(f)  7x2-9t/  =  G3. 

(g)  2x2-7  2/2  =  18. 
(h)  7x2 -2  2/2  =-8. 


2.  Transform  each  of  the  following  equations  by  translation  of  the 
axes  so  that  the  transformed  equation  shall  lack  terms  of  the  first  degree 
in  the  new  coordinates.   Draw  the  figure. 

(a)  4x2-2/2  +  8x-2  2/-l  =  0.  (d)  4x2  -  2/2  -  6x  -  4y  =  0. 

(b)  9x2-i/2  +  l8x-42/+  14  =  0.         (e)  x2  -  52/2  +  6x  -  lOy  =  0. 

(c)  3x2-2/2  +  12x  +  22/ +  14  =  0.         (f)  x^  -  2y^ +  10y  =  0. 

3.  Find  the  equations  of  the  following  hyperbolas  : 

(a)  transverse  axis  =  6,  foci  (—  2,  0)  and  (6,  0). 

Ans.    7  (x- 2)2 -9 2/3  =  63. 

(b)  conjugate  axis  =  6,  foci  (0,  2)  and  (0,  —  8). 

(c)  conjugate  axis  =  4,  foci  (1,  2)  and  (—  4,  2). 

(d)  transverse  axis  =  2,  foci  (0,  0)  and  (—  4,  0). 


170  NEW  ANALYTIC   GEOMETRY 

66.  Conjugate  hyperbolas  and  asymptotes.  Two  hyperbolas 
are  called  conjugate  hyperbolas  if  the  transverse  and  conjugate 
axes  of  one  are  respectively  the  conjugate  and  transverse  axes 
of  the  other. 

If  the  equation  of  a  hyperbola  is  given  in  typical  form, 
then  the  eqiiatlon  of  the  conjugate  hyperbola  Isfouyid  hij  cJuinging 
the  signs  of  the  coefficients  ofx^  and  if'  in  the  gloen  ef^uation. 

Thus  the  loci  of  the  equations 

(1)  1Q>  x"  -  7f  =10,   and    -1Q>  x"  +  if  =1Q 
are  conjugate  hyperbolas.    They  may  be  written 

1-1«=1    =">*    -1  +  16=1- 

The  foci  of  the  first  are  on  the  ic-axis,  those  of  the  second 
■on  the  ?/-axis.  The  transverse  axis  of  the  first  and  the  conju- 
gate axis  of  the  second  are  equal  to  2,  while  the  conjugate  axis 
■of  the  first  and  the  transverse  axis  of  the  second  are  equal  to  8. 

The  foci  of  two  conjugate  hyperbolas  are  equally  distant 
from  the  origin.  For  c^  equals  the  sum  of  the  squares  of  the 
semitransverse  and  semiconjugate  axes,  and  that  sum  is  the 
same  for  two  conjugate  hyperbolas. 

Thus  in  the  first  of  the  hyperbolas  above  c^  =1  +  16,  while 
in  the  second  c"^  =  16  + 1. 

If  in  one  of  the  typical  forms  of  the  equation  of  a  hyper- 
bola we  replace  the  constant  term  by  zero,  then  the  locus  of  the 
new  equation  is  a  pair  of  lines  (Theorem,  p.  40)  which  are 
called  the  asymptotes  of  the  hyperbola. 

Thus  the  asymptotes  of  the  hyperbola 

(2)  6V  -  ahf  =  a%'' 
are  the  lines 

(3)  ^»v-ay  =  o, 

or 

(4)  hdi-  -\-  ai/  =  0    and    bx  —  ay  =  0. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


171 


These  may  be  written 


(5) 


y  ^  —  —  X    and  7/  =  -  a;. 


They  pass  through  the  origin  and  their  slopes  are  respectively 

h       ,  I 
and  -• 

a  a 

The  property  of  these  lines  which  they  have  in  common  with. 
the  vertical  or  horizontal  asymptotes  of  Art.  22  is  expressed 
in  the 

Theorem.  The  branches  of  the  hijpevhola  ajjproach  indefinitehj 
near  its  asymptotes  as  they  recede  to  infinity. 

Proof.  Let  P-^(x^,  y^  be  a  point  on  either  branch  of  (2)  near 
the  asymptote  ^^  _  ^^  ^  0. 

The  perpendicular  distance  from  this  line  to  P^  is 
hx^  —  ay^  _  Yk 


(6) 


d  = 


—  -\llr  +  a^ 

We  may  find  a  value  for  the 

numerator  as  follows : 

Since  P^  lies  on  (2), 

Ip-xl  —  i^yl  =  a^lP-. 

Factoring  and  dividing, 

a^h"- 

bx^  —  ay.  = ; 

1         "^1       bx^  +  ay^ 

Substituting  in  (6),     d  . — 

—  V^'-^  +  a^  (bx^  +  ay^ 

As  Pj  recedes  to  infinity  in  the  first  quadrant,  x^  and  y^  be- 
come infinite  and  d  approaches  zero. 

Hence  the  curve  approaches  closer  and  closer  to  its  asymp- 
totes. Q.E.  D. 

Two  conjugate  hyperbolas  have  the  same  asymptotes. 

Thus  the  asymptotes  of  the  conjugate  hyperbolas  (1)  are  respectively 
the  loci  of  16x2-^2  =  0    and     -16x2  +  2/2  =  0, 

which  are  the  same. 


172 


NEW  ANALYTIC   GEOMETRY 


A  hyperbola  may  be  drawn  witb  fair  accuracy  by  the  fol- 
lowing 

Construction.  Lay  off  OA  =  OA'  =  a  on  the  axis  on  which  the 
foci  lie,  and  05  =  05'=  6  on  the  other  axis.  Draw  lines  through 
A,  A',  B,  B',  parallel  to  the  axes,  forming  a  rectangle.  Draw 
the  diagonals  of  the  rectangle.  Then  the  length  of  each  diago- 
nal is  obviously  2  c  (since  cv^  +  1/  =  c").  Moreover,  the  diagonals 
produced  are  the 
asymptotes.  For 
the  equations  of 
the  diagonals  ate 
readily  seen  to  be 
Itx  —  ai/  ^=  0  and 
hx  -\-  ay  =  0,  and 
these  are  the  same 
as  (4).  Construct 
the  circle  which 
circumscribes  the 
rectangle.  Draw  the  branches  of  the  hyperbola  tangent  to  the 
sides  of  the  rectangle  at  A  and  A'  and  approaching  nearer 
and  nearer  to  the  diagonals.  The  conjugate  hyperbola  may  be 
drawn  tangent  to  the  sides  of  the  rectangle  at  B  and  B'  and 
approaching  the  diagonals.  The  foci  of  both  are  the  points  in 
which  the  circle  cuts  the  axes. 

From  this  construction  the  influence  of  the  value  of  the 
eccentricity  upon  the  shape  of  the  hyperbola  can  be  easily  dis- 
cussed.   In  the  figure,  let  yl.4'  be  fixed.    Now  from  (12),  Art.  65, 

When  e  diminishes  towards  unity,  b  decreases,  the  altitude 
BB'  of  the  rectangle  diminishes,  the  asymptotes  turn  towards 
the  a;-axis,  and  the  hyperbola  flattens. 

When  e  increases,  the  asymptotes  turn  from  the  a;-axis,  and 
the  hyperbola  broadens. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


173 


67.  Equilateral  or  rectangular  hyperbola.  When  the  axes  of 
a  hyperbola  are  equal  (a  =  Ji),  the  hyperbola  is  said  to  be 
equilateral.  If  we  set  a  =  b\\\  equation 
(IV),  we  obtain 
(1)  x^  —  if=  a^, 

which  is  accordingly  the  equation  of 
an  equilateral  hyperbola  whose  trans- 
verse axis  lies  on  A'A''. 

Its  asymptotes  are  the  lines 
X  —  y  =  0    and   x  +  ?/  =  0. 

These  lines  are  perpendicular,  and  hence  they  may  be  used 
as  coordinate  axes.  The  designation  "  rectangular  "  hyj)erbola 
arises  from  this  fact. 

Theorem.    The  equation  of  an  equilateral  hyperbola  referred 
to  its  asymptotes  is 
(V)  %xy  =  d'. 

Proof  The  axes  must  be  rotated  through  —  45°  to  coincide 
with  the  asymptotes.    Hence  we  substitute  (Art.  55) 

x'  4-  ?/'  _  —  x'  +  y' 


V2 
m  (1).    This  gives 


y 


V2 


(-  x^  +  y'f 
2 


Reducing  and  dropping  primes  we  have  (V).  Q.e.d. 

It  is  important  to  observe  that  (V)  has  the  simple  form 
(2)  cfy  =  a  constant. 

68.  Construction  of  the" hyperbola.   A  mechanical  construction, 

depending  upon  the  definition  (1)  of  Art.  G5,  is  the  following : 
Fasten  thumb  tacks  at  the  foci.    Pass  over  F'  and  around  F 

a  string  whose  ends  are  held  together  (Fig.  1,  p.  174). 

If  a  pencil  be  tied  to  the  string  at  P,  and  both  strings  be 

pulled  in  or  let  out  together,  then  PF'  —  PF  will  be  constant 


174 


NEW  ANALYTIC   GEOMETRY 


and  P  will  describe  a  hyperbola.  If  the  transverse  axis  is  to  be  2  a, 
the  strings  must  be  adjusted  at  the  start  so  that  the  difference 
between  PF'  and  PF  equals  2  a. 

A  construction  often  used  for  an 
equilateral  hyperbola  when  the  asymp- 
totes and  one  point  A  are  given,  is  as 
follows  (Fig.  2) : 

Let  OX  and  0  Y  be  the  asymptotes 
and  A  the  given  point.  Draw  any  line 
through  A  to  meet  OX  at  M  and  O  F  at  iV. 

Lay  off  MP  =AN.  Then  P  is  a  point 
on  the  required  hyperbola. 

Proof.    Choose  the  asymptotes 
as  axes.    Let  the  coordinates  of  A 
be  {a,  h)  and  of  P,  (x,  y).    Then 
OS  =  x,  SP  =  y,  OB  =  1),  BA  =  a. 

By  construction,    AN  =  MP. 
,".  triangle  PSM  =  triangle  NBA, 
and  BN  =SP  =  y,  SM  =  AB  =  a. 

Since  the  triangles   03IN  and 
ABN  are  similar, 


BN  _  UN 

' '  Ib  ~~om 

Substituting, 
y_  ^J  +  y 

a 


OB  +  BN 
OS  -}-  sm' 


or    xy  =  ab. 


My  Mo 

Fig.  3 


a  +  ■*■ 

Comparing  with  (V),  we  see 
that  P  («,  y)  lies  upon  an  equilat- 
eral hyperbola  which  has  OX  and 
OY  for  its  asymptotes  and  which  passes  through  (a,h).     Q. e.d. 

By  drawing  different  lines  through  A,  and  laying  off 
-V^P^  =  AN^,  M^P^  =  AN,^,  etc.,  we  determine  as  many  points 
J\,  P.,,  etc.,  as  we  wish  on  the  hyperbola  (Fig.  3). 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  175 

PROBLEMS 

1.  Find  the  equations  of  the  asymptotes  and  of  the  hyperbolas  conju- 
gate to  the  following  hyperbolas,  and  plot : 

(a)  4  X-  -  y-^  =  36.  (c)  16  x^  -  y^  +  M  =  0. 

(b)  9x2-251/2  =  100.  (d)  8a;2-16?/2+ 25  =  0. 

2.  The  distance  from  an  asymptote  of  a  hyperbola  to  either  focus  is 
numerically  equal  to  5. 

>j  3.  The  distance  from  the  center  to  a  line  drawn  through  a  focus  of  a 
hyperbola  perpendicular  to  an  asymptote  is  numerically  equal  to  a. 

J  4.  The  product  of  the  distances  from  the  asymptotes  to  any  point  on 
the  hyperbola  is  constant. 

,•"   5.  The  focal  radius  of  a  point  P^  (x^,  y^)  on  the  parabola  2/2  —  2px  is 

2+^1- 

'      6.  The  ordinates  of  points  on  an  ellipse  and  the  major  auxiliary  circle 
which  have  the  same  abscissas  are  in  the  ratio  of  6 :  a. 

f    7.  The  area  of  an  ellipse  is  irab. 

Hint.  Divide  the  major  axis  into  equal  parts.  With  these  as  bases  inscribe 
rectangles  in  the  ellipse  and  major  auxiliary  circle  (p.  164).  Apply  Problem  6 
and  increase  the  number  of  rectangles  indefinitely. 

69.  The  examples  of  Problem  2,  p.  1G9,  illustrated  the  fact 
that  any  equation  of  the  second  degree  lacking  an  ajy-term,  but 
containing  a-'-^  and  ?/'^  with  coefficients  of  unlike  signs,  can  by 
translation  of  the  axes  be  transformed  into  the  form  (9) 

Ax^  +  Btj'^  =  a, 

in  which  .4  and  B  differ  in  sign. 

From  the  preceding  it  is  clear  that  the  locus  of  this  equation 
is  a  hyperbola  if  C  is  not  zero,  and  a  pair  of  intersecting 
lines  if  C  is  zero.    Hence  the 

Theorem.  If  an  equation  of  the  second  degree  contains  no 
xy-tevni,  and  If  x^  and  if  occur  ivlth  coefficients  differing  In  sign, 
the  locus  is  either  a  hyperbola  or  a  pair  of  intersecting  lines. 


176 


NEW  ANALYTIC   GEOMETRY 


70.  Locus  of  any  equation  of  the  second  degree.  The  locus 
problems  of  this  chapter  have  led  to  the  equations  of  the  sec- 
ond degree, 

(1)  y  =  2^;»cc    and   x^  =  '^py^ 

(2)  iP'x^  +  aSf  =  d-J}^   and    V'x^  —  a?y-  =  a^V^. 

These  are  simple  types,  of  course.  The  question  is,  however, 
this : 

Given  an  equation  of  the  second  degree,  can  the  equation  he 
transfoi'med  by  translating  and  rotating  the  axes  so  that  the 
transformed  equation  will  reduce  to  one  of  these  simple  types? 

To  answer  this  question,  take  the  general  equation  of  the  sec- 
ond degree,  namely, 

(3)  Ax^  +  Bxy  -}-  Cf  +  Dx  +  Ey  +  F  =  0. 

This  equation  contains  every  term  that  can  appear  in  an 
equation  of  the  second  degree. 

We  begin  by  rotating  the  axes  through  an  angle  6.  To 
do  this,  set  in  (3), 

ic  =  cc'  cos  d  —  y'  sin 0, 
and  y  =  ^'  sin  &  +  y'  cos  6. 

This  gives,  after  squaring,  multiplying,  and  collecting,  the 
transformed  equation 

(4)  A  cos"-' d  x''^-2  A  sin ^  cos  ^         x'y'  +  As\n^d 

+  5sin(9cos^         +5(cos-^— sin-6')  —  ^sin^cos^   y'^ 

+  C  sin"  ^  +  2  C  sin  ^  cos  6  +  C  cos-  0 

+  Dcos6  x'  —  DsinO   y' +  F  =  0. 

+  Esm6       -\-Ecose 

The  angle  $  is,  as  yet,  any  angle  at  all.    But  let  us  now,  if 

possible,  choose  this  angle  so  that  the  equation  (4)  shall  not 

contain  the  x'y'-teTva..   To  do  this,  we  must  set  the  coefficient 

of  x'y'  equal  to  zero ;  that  is, 

(5)  -2  A  sin  ecos6  +  B  (cos^^  -  sin^^)  +  2  C  sin  ^  cos  ^  =  0. 
But     2  sin  6  cos  6  —  sin  2  0,     cos'^O  —  sin'-^  =  cos  2  6. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  177 

Hence  (5)  becomes 

(6)  (C  -  ^)  sin  2  ^  +  i^  cos  2  6  =  0. 
Dividing  through  by  cos  2  6,  and  transposing, 

(7)  tan2^  =  j4-^- 

Since  any  number  may  be  the  tangent  of  an  angle,  it  is 
always  possible  to  find  a  value  for  6  from  this  equation.  If, 
then,  the  axes  are  rotated  through  the  angle  0  determined  by 
(7),  equation  (3)  reduces  to 

(8)  A  'a;'2  4-  C'lf^  +  D'x'  +  E'y'  +  F  =  0, 
where  from  (4), 

(9)  A'  =  A  cos'B  +  B  sin ^ cos ^  +  C  sin'^^, 

(10)  C'=  A  &\n^d  —  B  sin^  cos^  +  C  cos^B. 

The  discussion  gives  the 

Theorem.  The  term  in  xy  may  alivays  he  removed  from  an 
equation  of  the  second  degree, 

Ao-}  +  Bxy  -\-  Cy"^  -[-  Dx  -{-  Ey  -[-  F  =  0, 

by  rotating  the  axes  through  an  angle  6  such  that 

(VI)  ^"2^  =  73^- 

Now  equation  (8)  is  of  a  form  which  we  have  met  frequently 
in  this  chapter,  and  we  have  learned  to  simplify  it  by  transla- 
tion of  the  axes.  We  saw  in  Art.  61  that  if  only  one  square 
(yl'=  0,  or  C' =  0)  and  the  first  power  of  the  other  coordinate 
were  present,  the  equation  could  be  transformed  into  one  of  the 
typical  forms  (1)  of  the  parabola. 

Suppose,  however,  that  the  first  power  of  the  other  coordi- 
nate does  not  appear.  For  example,  suppose  in  (8)  that  /I '  =  0 
and  /)'  =  0.    Then  the  equation  is 


178  NEW  ANALYTIC  GEOMETRY 

This  is  an  ordinary  quadratic  in  y.  If  tlie  roots  are  real,  the 
locus  will  be  two  lines  parallel  to  the  cc'-axis.  These  lines  will 
coincide  if  the  roots  are  equal.  There  will  be  no  locus  if  the 
roots  are  imaginary. 

If  neither  A^  nor  C  is  zero,  we  may,  by  translation  to  the 

new  origin  {  —  7-—  >  —  -^-—^^  \  transform  the  equation  into 
\      Z  A  Z  C  I 

(11)  A'x'""  +  C'y"'^  +  F'  =  0. 

The  locus  of  this  equation  has  been  discussed  in  Arts.  64 
and  69. 

The  result  we  have  established  is  expressed  in  the 

Theorem.  The  locus  of  an  equation  of  the  second  degree  is 
either  a  parabola,  an  ellipse,  a  hyperbola,  two  straight  lines 
(which  may  coincide),  or  a  point. 

The  following  conclusion  also  may  be  drawn :  The  presence 
of  the  xy-term  indicates  that  the  axes  of  the  curve  are  not  paral- 
lel to  the  axes  of  coordinates. 

We  seek  now  a  test  to  apply  to  an  equation  containing  an 
icy-term  in  order  to  decide  in  advance  the  nature  of  the  locus. 
To  do  this  we  eliminate  the  angle  0  from  equations  (9)  and 
(10),  making  use  of  (6).    The  result  is  the  simple  equation, 

(VII)  -4:A'C'=B^-^AC. 

The  steps  in  the  ehmination  process  are  as  follows : 
Adding  and  subtracting  (9)  and  (10), 

(12)  A'+  C'  =  A  +  C  (since  sin2^  +  cos2^  =  1). 

(13)  A'- C'={A- C)cos20  +  Bsm2e. 
Squaring  (13), 

(14)  ( j^'_  C')2  =  {A-  Cf  cos2  2  6"  +  2  B (J-  -  C)  sin  2 ^i cos 2 (9  +  52  sin2  2 0. 

Squaring  (6), 
(16)      0  =  (4-  C)2sin22(9+  2  B(C  -  J.)sin2(9cos2^  + -K^cos22^. 

Adding  (14)  and  (15), 
(16)  {A'  -  C'f  =  {A-  C)2  +  B2. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  179 

Squaring  (12), 

(17)  {A'  +  cy  =  {A  +  Cf. 

Subtracting  (16)  and  (17),  we  obtain  (VII). 

If  the  locus  of  (8)  is  a  parabola,  .4 '  =  0  or  C  =  0.  Hence  from 
(VII),  £'-4^C  =  0. 

If  the  locus  of  (8)  is  an  ellipse,  A'  and  C  agree  in  sign. 
Hence  ^'C  is  positive,  and  from  (VII), ii^—  4:  AC  is  negative. 

If  the  locus  of  (8)  is  a  hyperbola,  A'  and  C"  differ  in  sign. 
Hence  A'C  is  negative,  and  from  (VII),  E'—  ^.AC  is  a  posi- 
tive number. 

Collecting  all  the  results  in  tabular  form,  we  have  the 

Theorem.     Given  any  equation  of  the  second  degree, 

Ax^  +  Bxy  +  Cif  -\- Dx  +  El/  -i- F  =  0. 
The  possible  loci  may  he  classified  thus: 


Test 

General  case 

Exceptional  cases  "■ 

zero 

parabola 

two  parallel  lines 
one  line 

B2_4^C 
negative 

ellipse 

point-ellipse 

m-AAc 

positive 

hyperbola 

two  intersecting  lines 

A  point-ellipse  is  often  called  a  "  degenerate  ellipse,"  two 
intersecting  lines  a  "  degenerate  hyperbola,"  and  two  parallel 
lines  a  "  degenerate  parabola." 

Note  that  B'^  —  A:  AC  is  the  discriminant  of  the  terms  of  the 
second  degree  in  the  equation. 

*For  tests  to  distinguisli  the  exceptional  cases,  see  Smith  and  Gale's  "  Ele- 
ments of  Analytic  Geometry,"  p.  277. 

t  This  case  is  recognizable  by  inspection,  for  the  terms  of  the  second  degree, 
Ax^  +  Bxy  +  Cy'^,  now  will  form  a  perfect  square. 


180 


XEW  AXALYTIC  GEOMETRY 


The  exceptional  cases  are  recognizable  by  the  condition  that 
the  equation  is  then  factorable  into  two  factors  of  the  first 
degree  in  x  and  y.  A  number  of  problems  of  this  kind  were 
given  on  page  41.  When  the  equation  is  not  readily  factored  by 
trial,  it  may  appear  by  the  first  method  of  the  following  section 
(Art.  71)  that  factors  do  nevertheless  exist.  Moreover,  under 
the  two  lirst  cases  in  the  table  (parabola  and  ellipse)  there  may 
be  no  locus.  This  fact  will  also  readily  appear  by  the  first 
method  of  Art.  71. 

71.  Plotting  the  locus  of  an  equation  of  the  second  degree.    In 

this    section    we    discuss    methods  of  plotting   second-degree 
equations  which  contain  a:-?/-terms. 

First  Method.  Bij  direct  jjlotting.  Test  by  the  theorem  at 
the  end  of  the  preceding  section,  and  then  plot  the  equation 
directly. 

EXAMPLES 
1.  Plot  the  locus  of 
(1)  x^ —  2xy  + -iy^ —  Ax  =  0. 

Solution.    Here         ^=1,     ii=-2,     C  =  4. 

.•.  ii-  —  4^(7  =  4  —  10  =  —  12  =  a  negative  number. 
Hence  the  locus  is  an  ellipse.  Y,- 


X 

y 

1 

J 

-f 

0,4 

0 

3  ±  Vfj 

1 

4 

2 

^ 

-1 

(2) 


Solve  the  equation  for  x  as  follows  : 

[Collecting  terms  in  x  and  completing  the  square.] 


^3) 


.-.  X  =  2/  +  2  ±V(2-2/){2  +  3?/). 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 


181 


Solving  also  for  y, 
(4)  y  =  \x±  iVa:(16-3x). 

From  the  radicals  in  (3)  and  (4)  we  see  that  (see  p.  49) 

y  may  have  values  from  —  |  to  2  inclusive  ; 

X  may  have  values  from  0  to  -*/  inclusive. 
Hence  the  ellipse  lies  within  the  rectangle 

y=-h   y  =  ^^   x  =  o,   X  =  if-. 

Points  on  the  locus  may  be  found  from  (3)  as  in  the  table. 

2.  Determine  the  locus  of 

5  x-  +  4  xy  —  ?/■-  +  24  X  —  6  ?/  —  5  =  0. 

Solution.  ^  =  5,  5  =  4,  G'=-l.  .-.  F^  -  4^C=:  16  +  20  =  36. 
Hence,  from  the  table  of  Art.  70,  we  may  expect  a  hyperbola  or  a  pair 
of  intersecting  lines. 

Solve  the  equation  for  y  as  follows  : 

y^  -(4x  -  6)2/  +  (2x  -  3)"  =  5x-  +  24x  -  5  +  (2x  -  3)^ 
=  9x2  +  12x  +  4  =(3x  +  2)-2. 
[Collecting  terms  in  y  and  completing  the  square.] 
...  2/_(2x-.3)  =  ±(3x  +  2). 

Hence  the  locus  is  the  intersecting  lines 

y  =  bx  —  1     and     y  =—  x—  b. 


PROBLEMS 

1.  Test  and  plot  the  following  equations  : 

(a)  x^  —  2xy  +  2/^  —  5x  =  0.        (c)  4,xy  +  4?/^  +  4?/  +  4  =  0. 

(b)  4x?/  + 4y2_2x  +  3  =  0.       (d)  2x2  +  4x2/ +  4?/'-2  +  2x  -  3  =  0. 


(e 
(f 
(g 
{^ 
(i 
(J 
(k 
(1 

(m; 

(o 


x2  +  2x2/  +  2^2^2x  +  22/-l  =  0. 

3x2-12x2/  +  92/2  +  8X-12  2/  +  5  =  0, 

5x2  — 12x^  +  92/2  +  8x-12  2/  +  3  =  0. 

x2  +  X2/  +  2/2  +  3  2/  =  0. 

x2  +  2xy  +  42/2  +  6^  =  0. 

4x2  +  4x2/  +  ?/  +  6x  -  9  =  0. 

3x2  —  2x2/  +  2/'^-4x-6  =  0. 

x2  —  2  xy  +  5  2/2  -  8  2/  =  0. 

x2  —  4x2/  +  42/2  +  4x  +  22/  =  0. 

3x2  +  4x2/  +  y2  — 2x  — 1  =  0. 

3x2  +  8x2/+4  2/2  +  2x  +  42/  =  0. 


182  NEW  ANALYTIC  GEOMETRY 

Second  Method.  By  transformation.  If  the  icy-terra  is 
lacking,  we  have  seen  that  the  equation  may  be  simplified 
by  translating  the  axes.  The  transformed  equation  is  then 
readily  plotted  on  the  new  axes. 

When  the  xy-texrci  is  present,  rotate  the  axes  through  the 
angle  d  given  by  (VI), 

(5)  tan2^  =  j4^. 

The  term  in  xy  will  then  disappear  and  further  simplification 
is  accomplished  by  translation.  ' 

To  rotate,  we  substitute 

(6)  X  =  x'  cos  0  —  ?/'sin  $,     y  =  x'  sin  0  -{-  y'  cos  0. 

We  find  sin  6  and  cos  6  as  follows.   First  compute  cos  26  from 

(7)  cos  2  ^  =  ±  —=!=■     (26  and  28,  p.  3) 
^  ^  Vl  +  tan-2^  ^ 

From  (5),  2  ^  must  lie  in  the  first  or  second  quadrant,  so  the 
sign  in  (7)  must  be  the  same  as  in  (5).  6  will  then  be  acute ; 
and  from  40,  p.  4,  we  have 


/I -cos  2^            .            /I  +  cos  2  ^ 
(8)         sine=+yj ,     cos^  =  +  ^^^^ 


EXAMPLES 

1.  Construct  and  discuss  the  locus  of 

(9)  x2  + 4x?/  + 4?/2  +  12x— 6?/ =  0. 

Solution.    Here  ^  =  1,     B  =  i,     C  =  4. 

.-.  B'^  —  i  AC  =  0,  and  the  locus  is  a  parabola. 
"Write  the  equation  (9)  in  the  form 

(10)  (x  + 2?/)2  +  12x-6?/  =  0. 
We  rotate  the  axes  through  an  angle  0,  such  that 

4  4 

tan2^  = = 

1-4  3 


Then  by  (7), 
and  by  (8), 

(11) 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA 

cos  2  ^  =  -  I, 


183 


sin^ 


--=    and    cos  ( 

V5 


1 


The  equations  for  rotating  tlie  axes  are  therefore 


y  = 


2x'  +  y' 


Substituting    in    tlie    equation 
(10),  we  obtain 
6 


V5 


0. 


r| 

fx' 

\, 

l,D 

\, 

/ 

\ 

s. 

"^ 

\ 

/ 

^ 

■v. 

\ 

h 

"^ 

^ 

> 

J 

/ 

^ 

u 

r 

__^ 

^ 

f- 

\ 

X 

/j 

\ 

\ 

k. 

/ 

^& 

Hence  the  locus  is  a   parabola 
3 

for    which    'p  — ,    and    whose 

V5 
focus  is  on  the  ?/'-axis. 

The  figure  shows  both  sets  of  axes,  the  parabola,  its  focus  and  directrix. 


The  axis  OX'  lias  the  slope  tan  0  = 


=  2,  from  (11).  Hence  to  draw 


0X\  simply  draw  a  line  through  the  origin  whose  slope  equals  2. 

In  the  new  coordinates  the  focus  is  the   point    /  0,  — ^  )  and  the 

,...,,,.,             3  \     2V5/ 

directrix  is  the  line  ?/  = =• 

2V5 

2.  Construct  the  locus  of 

5x2  _|.  6xy  +  5?/2  +  22x  —  6?/  +  21  =0. 
Solution.    Here  .4  =  5,     B  =  6,     C  =  5. 

.-.  52  _  4  j^(7  =  36  _  IQO  =  _  64  =  a  negative  number. 

Hence  the  locus  is  an  ellipse. 

We  rotate  the  axes  through  the  angle  ^,  given  by 

tan  2  (9 


5-5 

.-.  2^  =  90°,     (9*  =  46°. 

Hence  the  equations  of  the  transformation  are 

__      x'  —v'  x'  +  y' 


V2 


y 


V2 


*■'  n  4  =  C,  the  angle  d  always  equals  45°. 


184 


NEW  ANALYTIC  GEOMETRY 


1^' 

^'t 

^'' 

N 

/ 

-i- 

\ 

\ 

/ 

\ 

^ 

/ 

/ 

\ 

/ 

I 

\ 

\ 

X 

Nj 

\ 

A\ 

\ 

< 

A 

\\ 

\ 

/ 

y 

V 

\ 

^ 

/ 

/ 

X 

\) 

X 

/ 

\ 

Substituting  in  the  given  equation  and  reducing, 

4x'2  +  ?/'2  +  4 V2x'-  7  Vi/  +  -2/  =  0. 
Translating  to  the  new  origin  (— |\/2,  |V2),  the  final  equation  is 
4x"-^  +  ?/"2  =  16. 

Hence  the  locus  is  an  ellipse  whose 
major  axis  is  8,  whose  minor  axis  is  4,  -j^ 
and  whose  foci  are  on  the  F"-axis. 

The  figure  shows  the  three  sets  of 
axes  and  the  ellipse.  The  coordinates 
of  the  new  origin  0'(— |V2,  fVi) 
refer  to  the  axes  OX'  and  0Y\  and  this 
must  be  remembered  in  plotting. 

The  equation 

(12)  Bxij  +  Dx-^Ey  +  F  =^, 

in  which  a;^  and  if'  are  lacking,  offers  an  exception  to  the  above 
process,  for,  by  translation,  the  equation  may  be  reduced  to 

(13)  Bx'y'  +  F'=0; 

and  the  locus  of  (13)  is,  by  (V),  Art.  67,  an  equilateral  hyper- 
bola referred  to  its  asymptotes  as  axes.  Hence  to  plot  (12), 
translate  so  that  the  terms  of  the  first  degree  disappear  and 
then  plot  the  new  equation. 

To  show  that  (12)  may  be  transformed  into  (13)  by  translation,  proceed 
thus : 

Substitute  x  =  x'  +  /i,  ?/  =  ?/'  +  i,  in  (12),  multiply  out  and  collect  the 
terms.  We  obtain 


(14) 


Choose  the  new  origin  (A,  k)  so  that  the 

coefficient  of  x'  vanishes ;  that  is,  Bk  +  D  =  0, 

coefficient  of  y'  vanishes ;  that  is,  Bh  +  E  =  0. 

E  D 

— ,  k= ,  and  (14)  reduces  to  the 

B  B 


Bx'y'  +  Bk 

x'  +  Bh 

y'  +  Bhk 

+  B 

+  E 

+  Dh 
+  Ek 

+  F 

Solving  these  equations,  h  =  — 
form  (13). 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  185 

PROBLEMS 

1.  Simplify  the  following  equations  and  construct  the  loci.    Check  the 
figure  by  finding  the  intercepts  on  the  original  axes. 

V,      (a)  x^  +  xy  +  2j"  =  3. 

(b)  X2  +  3XT/  +  ?/  +  4y  =  o. 

(c)  x''  +  2xy  +  7/'  +  Sx  -  Sy  =  0. 

(d)  3x2 -4xv/  +  8x-l  =  0. 
(^     (e)  4x2  +  4x1/ +  2/'^  +  8x  — 16?/ =  0. 

(f)  3xy  +  4x  +  6y  +  l  =  0. 

(g)  17x2-12x?/  + 8?/2-68x  +  24?/-12 


Ans. 

3x'2  +  ?/'2:=6. 

Ans, 

25x"2-52/"2  +  32^0, 

Ans. 

2x'2-3V2?/  =  0. 

Ans. 

x"2_4  2/"2  +  l  =  0. 

>.     Ans. 

5x'"-8Vly'  =  0. 

Ans. 

3xy-  7  =  0. 

!/-12  = 

0- 

Ans. 

x"2  +  4?/'2-l6  =  0, 

Ans. 

?/'2  +  6x'=  0. 

Ans. 

6xy +  11  =  0. 

Ans. 

4x"2-92/"2  =  36. 

12  =  0. 

Ans. 

52  ?/'2  _  49  =  0. 

E/  +  43  = 

0. 

(h)  2/2  +  6x-  6?/  +  21  =  0. 
(i)  6xi/  + 4X-12  2/  + 3  =  0. 
( j )  12x2/  -  -5'/  +  48  ?/  -  30  =  0. 
(k)  4x2-12x?/  +  9?/2  + 2x-3?/ 

(1)  12x2  +  8x2/ +  18  2/2 +  48x  +  16t/  + 43 

Ans.  4x2 +  2 2/2  =  1. 

(m)  7  x2  +  50 X2/  +  7  2/2  =  50.  Ans.  16  x'2  -  9  ?/2  =  25. 

(n)  x2  +  3x2/ -32/2  + 6x  =  0.  ^ns.  21  x"2  _  49 2/"2  =  72. 

I    (o)  16x2-24x2/  + 9?/2-60x- 802/ +  400  =  0. 

J.ns.  2/"^  — 4x"  =  0. 
2.  Show  that  the  general  equation 

^x2  +  Bxy  +  Cy^  +  Dx  +  Ey  +  F=0 

may  be  simplified  by  translation  only,  so  that  the  new  equation  contains 
no  terms  of  the  first  degree  in  x  and  y,  if  the  coordinates  of  the  new 
origin  {h,  k)  satisfy  the  equations 

2Ah  +  Bk  +  D  =  0,     Bh  +  2  Ck  +  E  =  0. 

Ilence  show  that  the  new  origin  {h,  k)  is  the  center  of  the  locus,  unless 
B-  —  4AC  =  0.    In  the  latter  case  the  transformation  fails. 

72.  Conic  sections.  Historically,  the  parabola,  ellipse,  and 
hyperbola  were  discovered  as  plane  sections  of  a  right  circular 
cone.  Hence  the  generic  term  used  for  them,  — conic  sections 
or  conies. 

A  definition  often  used,  which  will  include  all  conic  sections, 
is  the  following :    When  a  point  P  moves  so  that  its  distances 


186 


NEW  ANALYTIC   GEOMETRY 


from  a  given  fixed  point  and  a  given  fixed  line  ewe  in  a  constant 
ratio,  the  locus  is  a  conic. 

The  given  fixed  line  is  called  the  directrix,  the  fixed  point 
the  focus,  and  the  number  representing  the  ratio  of  the  dis- 
tances of  P  from  the  focus  and  directrix  is  called  the  eccen- 
tricity. 

In  Problem  3,  p.  51,  we  found  the  equation  for  any  conic 
to  be 
(1)  (1  -  e")  x^  ^if-^-px^f^  0, 

if  e  is  the  eccentricity,  YY^  is  the  directrix,  and  (p,  0)  is  the 
focus.  Now  (1)  has  no  .T//-term.  Hence  we  see  at  once  by 
■comparison  with  our  previous  results  that  a  conic  is 

a  parabola  tvhen  e  =  1, 
an  ellipse  when  e  <.  1, 
a  hyperbola  when  e  >  1. 

Clearly,  when  e  =  1  the  definition  of  the  conic  agrees  with 
that  already  given  for  the  parabola. 

The  ellipse  and  hyperbola,  each  having  a  center,  are  called 
central  conies. 

Focus  and  eccentricity,  as  used  in 
this  section,  agree  with  these  terms 
as  already  introduced.  This  fact  is 
left  to  the  student  to  prove  in  the 
following  problems. 

The  equation  of  a  conic  in  polar  co- 
ordinates is  readily  found.  We  may 
show  that  if  the  pole  is  the  focus  and  the  polar  axis  the  principal 
axis  of  a  conic  section,  then  the  polar  equation  of  the  conic  is 
_         ep 


D 

E 

T(p,e) 

y 

9/ 

/' 

/\e 

II 

p' 

F 

M         J 

I) 

(2) 


e  cos  6 ' 


where  e  is  the  eccentricity  and  p   is   the   distance  from   the 
•directrix  to  the  focus. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  187 

For  let  P  be  any  point  on  the  conic.    Then,  by  definition, 
FP 
Ef  =  '- 

From  the  figure,  FP  =  p, 

and  EP  =  HM  ^p  -}-  pcosO. 

Substituting  these  values  of  FP  and  EP,  we  have 
— ^ =  .;        . 


J)  +  p  COS  0 


or,  solving  for  p,  p  =  z ;;•  Q.E.  D. 

'^  1  — ecos^ 


PROBLEMS 

1.  Simplify  (1),  p.  186,  by  translation  of  the  axes  when  e  ^zi  1.       22 

Ans.    (1  -  e2)  x2  +  2/2  -    ^^ 


l-e2 

2.  Show  that  in  a  central  conic  the  focus  coincides  with  the  focus 
already  adopted.  Hence  show  that  a  central  conic  has  two  directrices, 
one  associated  by  the  above  definition  with  each  focus. 

3.  Prove  that  e  in  Problem  1  agrees  with  e  as  defined  in  Arts.  62  and  65. 

4.  Prove  that  the  focal  radii  of  appoint  {x,  y)  on  the  ellipse  (HI), 
p.  161,  are  u  +  ex  and  a  —  ex. 

5.  Prove  that  the  focal  radii  of  a  point  on  the  hyperbola  (IV),  p.  167, 
are  ex—  a  and  ex  +  a. 

LOCUS  PROBLEMS 

It  is  expected  that  the  locus  in  each  problem  will  be  constructed  and 
discussed  after  its  equation  is  found. 

•  1.  The  base  of  a  triangle  is  fixed  in  length  and  position.    Find  the 
locus  of  the  opposite  vertex  if 

(a)  the  sum  of  the  other  sides  is  constant.  Ans.    An  ellipse. 

(b)  the  difference  of  the  other  sides  is  constant.     Ans.   A  hyperbola. 

(c)  one  base  angle  is  double  the  other.  Ans.   A  hyperbola. 

(d)  the  sum  of  the  base  angles  is  constant.  Ans.   A  circle. 

(e)  the  difference  of  the  base  angles  is  constant.     Ans.    A  conic. 

( f )  the  product  of  the  tangents  of  the  base  angles  is  constant. 

Ans.    A  conic. 


188 


NEW  ANALYTIC  GEOMETRY 


(g)  the  product  of  the  other  sides  is  equal  to  the  square  of  half  the 

base.  ^"^-   A  lemniscate  (Ex.  2,  p.  122). 

(h)  the  median  to  one  of  the  other  sides  is  constant.     Am.   A  circle. 

2.  Find  the  locus  of  a  point  the  sum  of  the  squares  of  whose  distances 
from  (a)  the  sides  of  a  square,  (b)  the  vertices  of  a  square,  is  constant. 

Ans.    A  circle  in  each  case. 

3.  Find  the  locus  of  a  point  such  that  the  ratio  of  its  distance  from  a 
lixed  point  Pj  (Xj,  y^)  to  its  distance  from  a  given  line  Ax  +  By  +  C  =  0 
is  equal  to  a  constant  k. 

Ans.    {A-^  +  £2  _  k^A'~) x^-2  k^ABxy  +  {A'  +  B^-  k^B^)  y^ 

-  2 {AH^  +  B^x^  +  k^A C)x-2 {A'^y^  +  B^y^  +  k^BC) y 
+  (xf  +  y{)  {A^  +  J32)  _  k^C'  =  0. 

4.  Find  the  locus  of  a  point  such  that  the  ratio  of  the  square  of  its 
distance  from  a  fixed  line  to  its  distance  from  a  fixed  point  equals  a 
constant  k. 

Ans.  x^  —  k^{x  —  p)^  —  k^y^  =  0  if  the  ^/-axis  is  the  fixed  line  and 
the  X-axis  passes  through  the  fixed  point,  p  being  the  distance  from 
the  line  to  the  point. 


Systems  of  conies.    When  an  equation  of  the  second  degree 
contains  one  arbitrary  constant,  the  locus  is  a  system  of  conies. 


PARABOLA,  ELLIPSE,  AND  HYPERBOLA  189 

EXAMPLE 

Discuss  the  system  represented  by  ■  -\ =  1. 

^o  —  tc     y  —  fc 

Solution.  When  k<9  the  locus  is  an  ellipse  whose  foci  are  (±  c,  0), 
where  c-  =  (25  —  A;)  —  (9  —  A;)  =  16.  When  9  <.A:  <  2.5  the  locus  is  an 
hyperbola  whose  foci  are  (±  c,  0),  where  c^  =  (25  —  k)  —  (9  —  fc)  =  16. 
When  k  >  25  there  is  no  locus.  Since  the  ellipses  and  hyperbolas  have 
the  same  foci  (±  4,  0),  they  are  called  confocal. 

In  the  figure  the  locus  is  plotted  for  fc  =—  56,  —  24,  0,  7,  9,  11,  16,  21, 
24,  25.  As  k  increases  and  approaches  9,  the  ellipses  flatten  out  and  finally 
degenerate  into  the  x-axis,  and  as  fc  decreases  and  approaches  9,  the  hyper- 
bolas flatten  out  and  degenerate  into  the  x-axis.  As  fc  increases  and 
approaches  25,  the  two  branches  of  the  hyperbolas  lie  closer  to  the  2/-axis, 
and  in  the  limit  they  coincide  with  the  y-nxis. 


PROBLEMS 

1.  Plot  the  following  systems  of  conies  and  show  that  the  conies  of 
each  system  belong  to  the  same  type.  Draw  enough  conies  so  that  the 
degenerate  conies  of  the  system  appear  as  limiting  cases. 

(a)^  +  '^'  =  fc.  (c)^-^'  =  fc. 

^  ^  16       9  ^  '  16       9 

(b)  2/2  =  2  fcx.  (d)  x^  =  2  fcy  -  6. 

2.  Plot  the  following  systems  of  conies  and  show  that  all  of  the  conies 
of  each  system  are  confocal.  Discuss  degenerate  cases  and  show  that  two 
conies  of  each  system  pass  through  every  point  in  the  plane. 

,  ,       x^  y-  ,  ,   .       x^  "2 

(a)    TTT— 7  +  l^TT-^  =  1-  (0    TT. 7  + 


16 -fc      .36 -fc  ^   '  64-fc      16 -fc 

(b)  y^  =  2kx  +  fc2.  (d)  x^=2ky  +  fc2. 

3.  Plot  and  discuss  the  systems  : 

(a)  16  (X  -  fc)2  +  9  y2  =  144.  (c)  {y  -  fc)2  =  4  x. 

(b)  xy  =  fc.  (d)  4  (X  -  fc)2  -  9  {y  -  kf  =  36. 

4.  Plot  the  following  systems  and  discuss  the  locus  as  fc  approaches 
zero  and  infinity : 

(X-fc)2         ^2^  (^_fc)2_^^ 

fc-2            36  ^  '        fc2            36 


CHAPTER  XI 


TANGENTS 


-^  73.  Equation  of  the  tangent.  A  tangent  to  a  curve  at  a  point 
P^  is  obtained  as  follows.  Take  a  second  point  P^  on  the  curve 
near  P^.  Draw  the  secant  through 
Pj  and  P^. 

Now  let  P^  move  along  the 
curve  toward  P^.  The  secant  will 
turn  around  P^.  The  limiting 
position  of  the  secant  when  P^ 
reaches  P^  is  called  the  tangent 
at  P^. 

We  wish  to  calculate  the  slope  of  the  tangent  at  a  point  on  a 
curve.  Let  the  coordinates  of  P^  be  (x^,  y^  and  of  P^  (a-^  +  li, 
y^  +  k).    Then  ^  ^  .  „  _        yt 


slo2Je  of  secant  P^P^ 


h 


To  find  the  slope  of  the  tangent,  we  begin  by  finding  a  value 
for  -  >  the  slope  of  the  secant,  as  in  the  following  example. 


EXAMPLE 

Find  the  slope  of  the  tangent  to  the  curve  C  :  Sy  =  x^a,t  any 
point  P.^(oc^,  7/j)  on  C  (see  figure  on  "page  191). 

Solution.  Let  P^(x^,  y^  and  Pjx.^+h,  y^+k)  be  two  points  on  C. 
Then  since  these  coordinates  must  satisfy  the  equation  of  C, 

(2)  8  2/x  =  ^T, 

and  8(y^  +  A-)=(a-^  +  /0'; 

or 

(3)  Sij^  +  8k  =  xl  +  3  x^h  +  3  xjv"  +  h\ 

190 


TANGENTS 


191 


Subtracting  (2)  from  (3),  we  obtain 

Sk  =  h(3x^  +  3a;jA  +  A'"); 
k  _Sx^  +  3 xji  +  /«-" 
h~  8 


Factoring, 
and  hence 


=  slope  of  secant  P^P^. 

Now  as  P^  approaches  P^,  h  and  k  approach  zero,  and  when 
the  secant  becomes  a  tangent  to  the  curve,  h  and  k  are  both 
equal  to  zero. 

Hence  the  slope  m  of  the  tangent  at  P^  will  be  obtained 
from  the  above  value  of  the  slope  of  the  secant,  namely, 

S-Tj-  +  3  xji  +  Ji- 
■  ~8  ' 

0  and  also  k  =  0,  if  k  appeared  in  the  expres- 

3.rf 
m  =  —z —    A  ns. 


by  setting  h 
sion.    Hence 


The  method  employed  in  this  example  is  general  and  may 
be  formulated  in  the  following 

Rule  to  determine  the  slojye  of  the 
tangent  to  a  curve  C  at  a  point  P^  on  C. 

First  step.   Let  P^(x^,  y^  and  Pjx^  -\-  h, 

y^  +  k)  he  two  points  on  C.    Substitute 

their  coordinates  in  the  eqtiation  of  C 

and  subtract. 

Second  step.    Find   a  value  for  —  j  the 

-^       h 

slope  of  the  secant  through  P^  and  P^. 

Third  step.  Find  the  limiting  value 
of  the  result  of  the  second  step  lohen 
h  and  k  approach  zero.    This  value  is  the  required  slope. 

Having  found  the  slope  of  the  tangent  at  P^,  its  equation  is 
found  at  once  by  the  point-slope  formula.  The  point  /'^  is  called 
the  point  of  contact. 


192 


^EW  ANALYTIC  GEOMETRY 


EXAMPLE 

Find  the  equation  of  the  tangect  to  the  circle 
x2  +  ?/2  =  r^ 
at  the  point  of  contact  (x^,  y^). 

Solution.    Let  Pj  (Xj,  y^)  and  P„  {x-^  +  h, 
y^  +  k)  be  two  points  on  the  circle  C. 

Then  these  coordinates  must  satisfy  the    x 
equation  of  the  circle.    Therefore 

(1)  x'l  +  2/2  =  r2, 

and         (x^  +  hy-  +  (?/i  +  k)'^  =  r^  ; 
or 

(2)  cc  2  +  2  x^/i  +  /i2  +  2/2  +  2  z/^A:  +  fc'^  =  f^. 

Subtracting  (1)  from  (2),  we  have 

2x.^h  + Jfi +  2y^k  +  k^  =  0. 
Transposing  and  factoring,  this  becomes 

k{2y^  +  k)=-h{2x^  +  h). 

k  2x,  +  h 

Whence  -  =  —  — -* 

h  2y^  +  k 

=  slope  of  the  secant  through  Pj  and  P^. 

Letting  P„  approach  P^,  h  and  k  approach  zero,  so  that  m,  the  slope  of 

the  tangent  at  Pj,  is 

^1 
m  = -• 

Vi 
The  equation  of  the  tangent  at  P^  is  then 

2/  -  2/i  =  -  —  (-c  -  a^i), 

or  XjX  +  2/,?/  =  xf  +  2/2. 

This  equation  may  be  simplified.    For  by  (1), 

so  that  the  required  equation  is 

x^x  +  y{y  =  r^.  Q-  E- 1*- 


TANGENTS  193 

Theorem.    The  equation  of  the  tangent  to  the  circle 

C  :  X-  -{- If  —  r^ 
at  the  point  of  contact  P.^(x^,  y^  is 
(I)  XiX  +  y^y  =  r^. 

The  point  to  be  observed  in  this  proof  is  this  : 
Always  simplifj  the  equation  of  the  tangent  by  making  use 
of  the  equation  obtained  when  x^  and  y^  are  substituted  for  x 
and  y  in  the  equation  of  the  given  curve. 

In  the  equation  (I)  the  point  of  contact  is  {x^,  y^,  while 
(ic,  y)  is  any  point  on  the  tangent. 

In  like  manner  we  may  prove  the  following 
Theorem.    The  equation  of  the  tangent  at  the  point  of  contact 
Pi  (•«!»  Vi)  io  the 

elliqjse  iV  +  a^'if  =  aHP-  is  b^x^x  +  a^y^^y  =  a^t^  ; 
hr/perbola  V^x^  —  ahf  =  crlr'  is  b^x^x  —  a'^y^^y  =  a^ti^ ; 
parahola  'if-=2px    is  y^y  =  p(x-\-x^. 

PROBLEMS 

1.  Find  the  equations  of  the  tangent  to  each  of  the  following  curves 
at  the  point  of  contact  (x^,  y-^  -. 

(a)x2  =  2/)y.  Ans.  x^x-p{y  +  y^). 

1/  (b)  x^  +  y^  —  2rx.  Ans.  x-^x  +  y{y  =  r(x  +  x^). 

]/(c)  ?/2  _  4a;  ^.  3,  Ans.  y^y  =  2x  +  2x^  +  S. 

(d)  xy  =  a^.  Ans.  x-^y  +  y-^x  =  2  a~. 

(e)  x^  +  xy  =  4.  Ans.  2XjX  +  x^y  +  y^x  =  8. 
(i)  x^  +  y^  +  Dx  +  Ey  +  F=0.  ^^                   ^ 

Ans.    XjX  +  y^y  +  -{x  +  x^)  + -{y  +  y^)  +  F  =  0. 

■'   {g)y  =  ^^-  -^'^-    3xf X  — 2/ +  2?/i  =  0. 

(h)  y^^x^. 

(i)  y  =  -4x2  +  Bx+  C.  (m)  xy^  +  a  =  0. 

( j )  Ax^  +  By^  +  Cx  =  0.  (n)  x^y  +  b  =  0. 

(k)  Ax^  +  By^  =  0.  (o)  xy^  +  a^x  -  a%  =  0. 

l\)  Axy  +  Bx  +  Cy  =  0.  (p)  y-{2a- x)  =  x^. 


194  NEW  ANALYTIC  GEOMETRY 

74.  Taking  next  any  equation  of  the  second  degree,  we  may 
prove  the 

Theorem.    The  equation  of  the  tangent  to  the  locus  of 

Ax^  +  Bxij  +  Cif  +  Dx  +  Ey  +  F  =  0 

at  the  point  of  contact  P^(x^,  y^  is 

Proof.  Let  P^  (j-^,  y^  and  P^  (a;^  +  7i,  y^  +  k)  be  two  points  on  the  conic. 
Then 

(1 )  ^xf  +  Bx^y^  +  Cy^  +  Dx^  +  %i  +  F  =  0  and 

A{x^  +  hf  ■\-B  (x^  +  h)  (y^  +  ^-)  +  C  {y^  +  A:)2  +  Z»  (x^  +  A) 

+  £(l/i  +  A:)  +  ^=0. 
Clearing  of  parentheses, 

(2)  Ax^  +  2^x^A  +  4/i2  ^  J5xi?/^  +  Bx^fc  +  By^h  +  BM 

+  C?/i-  +  2  Cj/^fc  +  Cfc2  +  i)Xj  +  Z»A  +  £'?/^  4-  £'4  +  F  =  0. 
Subtracting  (1)  from  (2), 

(3)  2  ^Xjft  +  Ah-  +  Bx^fc  +  By^h  +  B/ifc  +  2  Cy^k  +  CA:2  +  DA  +  ^-fc  =  0. 
Transposing  all  the  terms  containing  A,  and  factoring,  (3)  becomes 

k{Bx^  +  2Cy^-\-  Ck+  E)=-h(2Ax^  +  Ah  +  By^  +  Bfc  +  D) ; 

k  2  Ax,  +  By,  +  D  +  Ah  +  Bk 

whence  -= 

h  Bx^  +  2Cy^  +  E+  Ck 

This  is  the  slope  of  the  secant  P^Po- 

Letting  P.,  approach  Pj,  h  and  k  will  approach  zero  and  the  slope  of 

the  tangent  is  2Ax,  +  By,  +  B 

m= 

Bx^  +  2  C2/1  +  £ 

The  equation  of  the  tangent  line  is  then 

2  Ax,  +  By,  +  D, 

y-y,  = ^-^ — ^^LZ —  (x  —  x,). 

•"'  Bx^  +  2Cy^  +  E^  '' 

To  reduce  this  equation  to  the  required  form  we  first  clear  of  fractions 
and  transpose.    This  gives 

{2AXj^  +  By^  +  B)x+  {Bx^  +  2Cy^  +  E)  y 

-  (2  Axl  +  2  Bx^y^  +  2  Cy^  +  Dx^+  Ey^)  =  0. 


TANGENTS  195 

But  from  (1)  the  last  parenthesis  in  this  equation  equals 

-{Dx^-\-Ey^  +  2F). 

Substituting,  the  equation  of  the  tangent  line  is 

{2Ax^  +  By^  +  D)x  +  (Bx^  +  2  Cy^  +  E)y  +  {Dx^  +  Ey^  +  2F)  =  0. 

Removing  the  parentheses,  collecting  the  coefficients  of  ^,  B,  C,  D,  E, 
and  F,  and  dividing  by  (2),  we  obtain  the  eqi;ation  of  the  theorem.    Q.  E.  D. 

The  above  result  enables  us  to  write  down  the  equation  of 
the  tangent  to  the  locus  of  any  equation  of  the  second  degree. 
For  by  comparing  the  equation  of  the  ciu've  and  the  equation 
of  the  tangent  we  obtain  the  following 

Rule  to  irrlte  the  equation  of  the  tangent  at  the  point  of  con- 
tact P^(x^,  ?/j)  to  the  locus  of  an  equation  of  the  second  degree. 

?/  IT  ~\~  K*  II 

Sid)stitute  x^x  and  y^j  for  x^  and  y'^,  — — - — —  for  xy,  and 
— —  and  — TT'^  Jo>'  •"'  «^«  y  i^n  ine  given  equation. 

For  example,  the  equation  of  the  tangent  at  the  point  of  contact  (x^,  ^,) 
to  the  conic  x-  +  3  x?/  —  4  2/  +  5  =  0  is 

x^x  4-  \  (x^2/  +  2/iX)  _!(?/  +  2/^)  +  5  =  0 ; 
or,  also,  (2  x j  +  3  y,)  x  +  (3  Xj  -  4)  ?/  -  4  j/j  +  10  =  0. 

75.  Equation  of  the  normal.  The  normal  to  a  curve  at  a  point  P^ 
is  the  line  drawn  through  P^  perpendicular  to  the  tangent  at  P  . 
When  the  equation  of  the  tangent  has  been  found,  we  may 
find  at  once  the  e(iuatiou  of  the  normal  in  the  manner  of  Chapi- 
ter IV.  Thus,  using  the  equations  of. the  tangents  given  on 
page  193,  we  find  the 

Theorem.    The  equation  of  the  normal  at  P^(x^,y^  to  the 
ellipse  b'^x^  +  «"y  =  a%'^      is    c^y^x  —  H^x^y  —  (d^  —  b"^)  x^y^ ; 

hyperbola      Ip-x^  —  a'^if  =  a-lr      is     d^y^x  -f-  b^x^y  =(0^  -\-b^)  x^^y^ ; 
parabola  y^  =  2  px    is  y^x  -\-  py  —  x^y^  -}-  py^. 


196 


NEW  ANALYTIC  GEOMETRY 


For  example,  for  the  ellipse  : 
The  slope  of  the  tangent 


A 

IS   m  = 

B 


b% 


Hence  the  equation  of  the  normal  is 

2/  -  2/1  =  ~  (-c  -  Xi), 
b-x. 


and  this  reduces  to  the  equation  in  the  tli^rem. 

In  numerical  examples  the  student  should  use  the  Rule  given 
to  write  down  the  equation  of  the  tangent,  find  the  normal  as 
a  perpendicular  line,  and  not  use  the  special  formulas. 

76.  Subtangent  and  subnormal.  If  the  tangent  and  normal  at 
P^  intersect  the  .r-axis  in  T  and  X  respectively,  then  we  define 

P^  T  =  length  of  tangent  at  P^,     Y' 


(1) 


P,N  =  leny-th  of  normal  at  P, 


The  projections  on  A'A'' of  P^Tand 
P^N  are  called  respectively  the  sub- 
tangent  and  s\(hnorvial  at  1^^.  That 
is,  in  the  figure, 

]\I^T  =  suhtangmt  at  P^, 
M^N  ^  subnonnal  at  P,. 


(2) 


The  subtangent  and  subnormal  are  readily  found  when  the 
equations  of  the  tangent  and  normal  are  known.  For,  from  the 
figure. 


(3) 
and 


M^T  =  0T 


OM,. 


M^N  =  ON  —  OM^ 
OM,  =  x„ 


while  OT  and  ON  are  respectively  the  intercepts  on  A'A'  of  the 
tangent  and  normal  at  7*^^.  Since  the  subtangent  and  sub- 
normal are  measured  in  opposite  directions  from  the  foot  of 
the  ordinate  M^P^,  they  will  have  opposite  signs. 


TANGENTS 


197 


EXAMPLE 

Find  the  equations  of  tangent  and  normal,  and  the  lengths  of  subtan- 
gent  and  subnormal  at  the  point  on  the  parabola  x^  =  4  y  whose  abscissa 
equals  3 

Solution.     The    point    of    contact 
(x,,  y.)  is 

The   formula   for   the    tagigent   at 
(Xj,  2/i)  is,  by  the  Rule,  p.  195, 
XjX  =  2{y  +  2/j). 
Substituting  the  values  of  a:^  and  ?/^, 
3x  =  2(i/  +  I)  or  6x  — 4?/-  9  =  0. 

This  is  the  required  equation  of  the 
tangent. 

The  slope  of  this  line  is  |.    Hence  the  equation  of  normal  at  (3,  |)  is 

2/- |=- t(-c-3),    or   8X  +  122/-51  =  0, 

TJie  intercept  on  A' A"' of  the  tangent  Is  f  ;  of  the  normal  \i.  Also  x^  =  3. 

.-.  subtangent  =  |  —  3  =  —  |, 

and  subnormal  =  y  —  3  =  --f-. 

The  lengths  of  the  tangents  and  normals  may  be  found  by  geometry, 
for  the  lengths  of  the  legs  of  the  triangles  P^M^  T  and  P^M^N  are  now 
known. 

PROBLEMS 

1.  Find  the  equations  of  the  tangent  and  normal  at  the  point  indicated 
to  each  of  the  following.  Find  also  the  lengths  of  subtangent  and  subnor- 
mal.  Draw  a  figure  in  each  case. 

(a)  2x2  +  3  2/2  =  35^  x^  =  2,  y^  positive.* 

Ans.   Tangent,  4  x  +  9  ?/  =  35 ;  normal,  9  x  —  4  y  =  6. 
Subtangent  =  \'- ;  subnormal  =—  |. 

(b)  x2  —  4  ?/2  +  15  =  0,  x^  =:  1,  y^  negative, 

(c)  2/2  =  4x  +  3,  yy-2. 

(d)  xy  =  4,  Xj  =  2. 

(e)  x2-f-2/2-4x-3  =  0,  Xi  =  3. 

(f)  x2  +  4?/2  +  5x  =  0,2/1  =  1. 

(g)  4x2  +  3y2  =  1 ;  positive  extremity  of  latus  rectum. 

*  Substituting  a;  =  2  in  the  given  equation,  we  find  y  =  ±  3.   Hence  ?/,  =+3. 


198 


NEW  ANALYTIC  GEOMETRY 


(h )  x2  +  x?/  +  4  =  0,  Xj  =  2. 

(i)  2/2  +  2a;2/-3  =  0,  2/j=-l. 

( j )  x2  —  3  x?/  —  4 1/2  +  9  =  0,  Xj  positive,  y^  =  2. 

(k)  x2  +  xy  +  ?/2  =  4,  x^  =  0,  j/j^  negative, 

( 1  )  x2  +  42/2  +  4x  -  82/ =  0,  x^  =  0. 

(m)  4  2/  =  x^,  x^  =  2. 

(n)  4^/2  =  x^,  x^  =  2. 

2.  Show  that  the  subtangent  in  the  parabola  y^  =  2j3x  is  bisected  at  the 
vertex,  and  that  the  subnormal  is  constant  and  equals  p. 

■/  77.    Tangent  whose  slope  is  given.    Let  it  be  required  to  find 
the  equation  of  a  tangent  to  the  ellipse 

(1)  5x^  +  2/^  =  5 
whose  slope  equals  2. 

Solution.  Draw  the  system  of  lines  whose  slope  equals  2 
(Art.  36).  We  observe  that  some  of  the  lines  intersect  the 
ellipse  in  two  points,  and  also  that  some  of  them  do  not  inter- 
sect the  ellipse  at  all.  Furthermore,  two  of  them  are  tangent. 
We  wish  to  find  the  equations  of 
these  two  tangents. 

The  equation  of  the  system  of 
lines  whose  slope  equals  2  is 

(2)  y  =  2x  +  k, 
where  k  is  an  arbitrary  parameter. 

Let  us  now  start  to  solve  for  the 
points  of  intersection.  Substituting 
from  (2)  into  (1), 

(3)  5  x'  +  {2x  +  kf  =  5. 
Squaring  and  collecting  terms, 

(4)  9  .T^  -H  4  kx  +  7r  -5  =  0. 

If  the  line  (2)  is  the  tangent  AB  of  the  figure,  by  solving 
equation  (4)  we  shall  obtain  the  abscissa  of  the  point  of  con- 
tact. But  (4)  is  a  quadratic  and  has  two  roots.  Hence  these 
roots  must  be  equal. 


TANGENTS  199 

We  learn  in  algebra  that  the  roots  of  the  quadratic 

(5)  Ax-  -{-Bx+C  =  0 
are  equal  when 

(6)  B''-4:AC  =  0. 

Comparing  (4)  with  (5), 

A  =  9,     B  =  4:k,     C  =  k''~5. 
Substituting  in  (6), 

(7)  16F-36(P_5^^^0,     or    A;  =  ±  3. 
Hence  the  equations  of  the  required  tangents  are 

AB:y=2x-^3    and     CD:y  =  2x-Z. 
Check.    AVriting  A-  =  3  in  (4),  it  becomes 

9  ic-  +  12  a:  +  4  =  0,     or    (3  a;  +  2f  =  0. 

The  equation  is  now  a  'perfect  square,  and  this  fact  consti- 
tutes the  check  desired.  Hence  the  equal  roots  have  the  com- 
mon value  ic  =  —  §.  This  is  the  abscissa  of  the  point  of  contact 
/'.  The  ordinate  is  found  from  y  =  2  a;  +  3  to  be  ^  =  |.  Hence 
J'  is  (-  f ,  I). 

Similarly,  putting  A-  =  —  3  in  (4),  we  find  Q  to  be  (§,  —  §). 

The  method  followed  in  the  preceding  may  be  thus  outlined. 

To  find  the  equation  of  the  tangent  to  a  conic  when  the  slope 
of  the  tangent  is  given. 

1.  Write  down  the  equation  of  the  system  of  lines  with  the 
given  slope  (y  =  mx  -f-  A:).  This  equation  contains  a  parameter 
(7j)  whose  value  must  be  found. 

2.  Eliminate  x  ox  y  from  the  equations  of  the  line  and  conic 
and  arrange  the  result  in  the  form  of  a  quadratic 

(8)  A^f  +  By-l^  C  =  0,     or     Ax^  +  Bx  +  C  =  0. 


200  NEW  ANALYTIC  GEOMETRY 

3.  The  roots  of  this  quadratic  must  be  equal.    Hence  set 
(9)  B'^-4.AC  =  0, 

and  solve  this  for  the  parameter  k. 

4.  Substitute  the  values  of  the  parameter  k  in  the  equation 
of  the  system  of  lines. 

5.  Check.  When  each  value  of  the  parameter  satisfying  (9) 
is  substituted  in  (8),  the  quadratic  becomes  a  perfect  square. 

PROBLEMS 

1.  Find  the  equations  of  tlie  tangents  to  tlie  following  conies  which 
satisfy  the  condition  indicated,  check,  and  find  the  points  of  contact. 
Verify  by  constructing  the  figure. 

(a)  7/2  =  4  X,  slope  =  |.  Ans.    x  —  2y  +  4:  =  0. 

(b)  x2  +  ?/2  =  16,  slope  =  -  |.  Ans.   ix  +  Sy  ±20  =  0. 

(c)  9  x2  +  16  2/2  =  144,  slope  =  -  i.  Ans.   x  +  4  ?/  ±  4  VlO  =  0. 

(d)  X-  —  4  ?/2  =  36,  perpendicular  to6x  —  4?/  +  9  =  0. 

Ans.    2x  +  3y  ±3\^  =  0. 
_„(e)  x2  +  2y2_x +  ?/  =  0,  slope  =  —  1.     Ans.  x  +  ?/  =  l,  2x  +  2?/  + 1  =  0. 

(f )  xy  +  y'^  —  4 X  +  8  y  =  0,  parallel  to2x  —  4y  =  7. 

Ans.    x  =  2y,x  —  2y  +  48  =  0. 

(g)  x^  +  2xy  +  y^  +  8x  —  Gy  =  0,  slope  =:  |.  Ans.  ix  —  Sy  =  0. 
(h)  x2 -f  2x?/  — 4x  +  2?/  =  0,  slope  =  2.  Ans.  y  =  2x,2x  —  y  +  10=0. 
(i )  2  x2  +  3  2/2  =  35,  slope  =  |.  ( 1 )  2/2  +  4  x  -  9  =  0,  slope  =  -  1. 
( j )  x2  +  2/^  =  2.5,  slope  =  —  |.            (m)  x^  —  y^  =  16,  slope  =  f . 

(k)  x^  +  4  2/  —  8  =  0,  slope  =  2.         ( n )  xy  —  4  =  0,  slope  =  —  |. 

78.  Formulas  for  tangents  when  the  slope  is  given.    For  later 

reference  we  collect  in  this  section  formulas  giving  the  equa- 
tions of  tangents  to  the  conies  in  terms  of  the  slope  m  of  the 
tangent.  The  student  should  derive  these  formulas,  following 
the  method  of  the  preceding  section. 

Theorem.  The  equation  of  a  tangent  in  terms  of  Its  .flops  m  to  the 


/  circle  x^  -\-  y^  =  r^       is  y  =  mx±r  Vl  + 


m'' 


/  ellipse         ly^x^  +  ahj'^  =  cc^lp-    is  y=mx±  ^a^m^  -\-  b"^ ; 

hyperbola   IP-x^  —  ahf  —  a%^    is  y  =  mxd:  ^a^m^  —  b^ ; 

P 
parabola  y  =  2px  is  y  =  mx  -\- 


TANGENTS 


201 


79.  Properties  of  tangents  and  normals  to  conies.  If  we  draw 
the  tangent  AB  and  the  normal  CD  at  any  point  P^  on  the 
ellipse,  and  if  we  draw  also  the 
focal  radii  P^F  and  P^F',  we  may 

prove  the  property  : 

The  tanjent  (ind  normal  to  an 
cUlpse  bisect  respectively  the  exter- 
nal and  internal  angles  formed 
hi/  the  focal  radii  of  the  point  of 
contact. 

Proof  In  the  figure  we  wish  to 
l)rove  6=  cf).   To  do  this  we  find  tan  (f>  and  tan  0  by  (VI),  Art.  35. 

The  slopes  of  the  lines  joining  P^  (x^,  y^  on  the  ellipse 

to  the  foci  F'  {c,  0)  and  F  (-  c,  0)  are 

slope  of  F'P^  =      -'^     ; 


slope  of  FP^  = 


x^-\-  c 


The  equation  of  the  tangent  AB  is  (Theorem,  Art.  73) 
b'^x^x  +  a'^y^y  =  a'b^. 


Now  tan  0  = 


b^x 
slope  of  AB  — ~  • 

J  where  m^  =  slope  of  AB,  m,^  =  slope 


ofP^F'.  1  +  ^1^2 

Substituting  the  above  values  of  the  slopes, 


tan  6 


bh'^ 

JLl 

«7/i 

X 

I  ~ 

c 

= 

-  b^x 

f  +  Wcx^ 

-  « V 

1 

U% 

'/l 

-  ^''^^Ji 

ahj 

i(-^i 

— 

') 

{"'iff 

+  b-'x^)  - 

-  b^cx^ 

aVi  - 

-  (a'  -  U 

)^-iZ/, 

202  NEW  ANALYTIC  GEOMETRY 

But  since  I\  lies  on  the  ellipse, 


a 


'■yl  -\-  U'Xi  =  a%'^ 


and  also  a^  —  V^  =  c^. 

a%^  —  J?cx.        Iria^  —  cx^        V' 

Hence       tan  6  =  —„ ;; =       ,  ., ~  =  —  • 

a'cij^  —  G-x^y^       cy^itr—cx^       cy^ 

In  like  manner, 

_      —  Irx^  —  Jrcxy  —  ci^yf     _  (li^x^  +  ft^yf )  +  b'^cx-^ 
—  «"^',3/i  —  «%i  +  ^'^iV^        «%i  +  («^  —  ^^)  x^y^ 

ahy^  +  c\yi       cy^ 

Hence  tan  0  =  tan  <^ ;  and  since  0  and  </>  are  both  less  than 
-TT,  0  =  <j>.  That  is,  AB  bisects  the  external  angle  of  FP^  and 
F'P^,  and  hence,  also,  CD  bisects  the  internal  angle.         Q.  E.  D. 

An  obvious  application  of  this  theorem  is  to  the  problem : 

To  draw  a  tangent  and  normal  at  a  g'loen  point  P^  on  an  ellipse. 

This  is  accomplished  hy  connecting  P^  with  the  foci  and 
bisecting  the  internal  and  external  angles  formed  by  these  lines. 

The  phenomenon  observed  in  "  whispering  galleries  "  depends 
upon  this  property;  namely,  let  the  elliptic  arc  A' PA  be  a 
vertical  section  of  such  a  gallery. 
The  waves  of  sound  from  a  person's 
voice  at  the  focus  F  will,  after 
meeting  the  ceiling  of  the  gallery, 
be  reflected  in   the    direction   F'. 

For  if  PN  is  the  normal  at  P,  angle  NPF  =  angle  NPF',  and 
the  law  of  reflection  of  sound  waves  is  precisely  that  the  angles 
of  incidence  (=  Z.  NPF)  and  reflection  (=  Z.  XPF')  are  equal. 
Hence  sound  waves  emanating  from  F  in  all  directions  will 
converge  at  F'.  A  whisper  at  F,  which  would  not  carry  over 
the  distance  FF',  might  consequently,  through  reflection,  be 
audible  at  F'. 


TANGENTS 


203 


In  like  manner  we  prove  the  following  properties  : 

The  tangent  and  normal  to  a  hyperbola  bisect  respectively  the 
internal  and  external  angles  formed  by  the  focal  radii  of  the 
point  of  contact. 

The  tangent  and  normal  to  a  parabola  bisect  respectively  the 
internal  and  external  angles  formed  by  the  focal  radius  of  the 
point  of  contact  and  the  line  through  that  point  parallel  to  the  cuxis. 


These  theorems  give  rules  for  constructing  the  tangent  and 
normal  to  these  conies  by  means  of  ruler  and  compasses. 

Construction.  To  construct  the  tangent  and  normal  to  a  hyper- 
bola at  any  point,  join  that  point  to  the  foci  and  bisect  the  angles 
formed  by  these  lines.  To  construct  the  tangent  and  normal  to 
a  parabola  at  any  point,  draw  lines  through  it  to  the  focus  and 
parallel  to  the  axis,  and  bisect  the  angles  formed  by  these  lines. 

The  principle  of  parabolic  reflectors  depends  upon  the  prop- 
erty of  tangent  and  normal  just  enunciated  ;  namely,  the  reflect- 
ing surface  of  such  a  reflector  is  obtained  by  revolving  a  para- 
bolic arc  about  its  axis.  If,  now,  a  light  be  placed  at  the  focus, 
the  rays  of  light  which  meet  the  surface  of  the  reflector  will  all 
be  reflected  in  the  direction  of  the  axis  of  the  parabola ;  for  a 
ray  meeting  the  surface  at  P^  in  the  figure  will  be  reflected  in  a 
direction  making  with  the  normal  PD  an  angle  equal  to  the  angle 
FP^D.  But  this  direction  is,  by  the  above  property,  parallel  to 
the  axis  OX  of  the  parabola. 


204  NEW  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Tangents  to  an  ellipse  and  its  major  auxiliary  circle  (p.  164)  at 
points  with  the  same  abscissa  intersect  on  the  x-axis. 

2.  The  point  of  contact  of  a  tangent  to  a  hyperbola  is  midway  be- 
tween the  points  in  which  the  tangent  meets  the  asymptotes. 

3.  The  foot  of  the  perpendicular  from  the  focus  of  a  parabola  to  a 
tangent  lies  on  the  tangent  at  the  vertex. 

4.  The  foot  of  the  perpendicular  from  a  focus  of  an  ellipse  to  a  tangent 
lies  on  the  major  auxiliary  circle  (p.  164). 

5.  Tangents  to  a  parabola  from  a  point  on  the  directrix  are  perpen- 
dicular to  each  other. 

,  6.  Tangents  to  a  parabola  at  the  extremities  of  a  chord  which  passes 
through  the  focus  are  perpendicular  to  each  other. 

7.  The  ordinate  of  the  point  of  intersection  of  the  directrix  of  a  parab- 
ola and  the  line  through  the  focus  perpendicular  to  a  tangent  is  the  same 
as  that  of  the  point  of  contact. 

8.  How  may  Problem  7  be  used  to  draw  a  tangent  to  a  parabola  ? 

9.  The  line  drawn  perpendicular  to  a  tangent  to  a  central  conic  from 
a  focus,  and  the  line  passing  through  the  center  and  the  point  of  contact 
intersect  on  the  corresponding  directrix  (Art.  72). 

10.  The  angle  which  one  tangent  to  a  parabola  makes  with  a  second  is 
half  the  angle  which  the  focal  radius  drawn  to  the  point  of  contact  of 
the  first  makes  with  that  drawn  to  the  point  of  contact  of  the  second. 

11.  The  product  of  the  distances  from  a  tangent  to  a  central  conic  to 
the  foci  is  constant. 

12.  Tangents  to  any  conic  at  the  ends  of  the  latus  rectum  pass  through 
the  intersection  of  the  directrix  and  principal  axis. 

13.  Tangents  to  a  parabola  at  the  extremities  of  the  latus  rectum  are 
perpendicular. 

14.  The  equation  of  the  parabola  referred  to  the  tangents  in  Problem 
13  IS  x2  -  2  xi/  +  2/2  -  2  y/2p  {x  +  y)  +  2p2  -  o. 

Show  that  this  equation  has  the  form  x-  +  y-  =  vpV2. 

15.  The  area  of  the  triangle  formed  by  a  tangent  to  a  hyperbola  and 
the  asymptotes  is  constant. 

16.  An  ellipse  and  a  hyperbola  which  are  confocal  intersect  at  right 
angles. 


CHAPTER  XII 


PARAMETRIC  EQUATIONS  AND  LOCI 


80.  If  X  and  y  are  rectangular  coordinates,  and  if  each  is  ex- 
pressed as  a  function  of  a  variable  parameter,  as,  for  example, 

(1)  x  =  \t\     y  =  \t\ 

in  whiQh  Hs  a  variable,  then  these  equations  are  called  the  jjara- 
metr'ic  equations  of  the  curve,  —  the  locus  of  (x,  y). 

To  plot  the  curve,  give  values  to  t  and  compute  values  of  x 
and  y,  arranging  the  work  in  a  table.  When  the  computation  is 
finished,  plot  the  points  (.r,  y)  and  draw  a  smooth  curve  through 
them. 

EXAMPLES 
1.  Plot  the  curve  whose  parametric  equations  are 

(2)  x  =  \r\     y^itfi-  _jYA 


t 

X 

y 

0 

0 

0 

1 

.5 

.25 

2 

2 

2 

3 

4.5 

0.75 

etc. 

etc. 

etc. 

-  1 

.5 

-   .25 

-2 

2 

-2 

-8 

4.5 

-  (S.75 

etc. 

etc. 

etc. 

Solution.   The  table  is  easily  made.    For  example,  if  i  =  2,  then  x  =  2, 
y  =  2,  etc. 

The  curve  is  called  a  semicubical  parabola. 

205 


206 


NEW  ANALYTIC   GEOMETRY 


2.  Draw  the  locus  of  the  equa- 
tions 

(3)      a;  =  2rcos^  +  rcos2^, 

?/  =  2  ?•  sin  9  —  r  sin  2  6, 

where  ^  is  a  variable  parameter. 

Solution.  Take  r  =  5.  Arrange 
the  computation  as  below  : 

The  three-pointed  curve  thus 
obtained  is  called  a  hypocycloid  of 
three  cusps. 


j:=  10  cos  6-1- 5  cos  2  0,    i/=  10  sin  9  — 5  sin  2  0 

6 

cosO 

20 

cos  2  0 

X 

sin0 

sin  2  0 

U» 

0 

1 

0 

1 

15 

0 

0       1 

C 

30° 

.86 

60° 

.50 

11.1 

.50 

.86 

0.7 

60° 

.50 

120° 

-.50 

2.5 

.86 

.86 

4.3 

90° 

0 

180° 

-1 

-6 

1 

0     ^^ 

120° 

-.50 

240° 

-.50 

—  7.5 

.86 

-.86 

12.9 

150° 

-.86 

300° 

.50 

-6.1 

.50 

-.86 

9.3 

180° 

-1 

360° 

1 

-5 

0 

"4 

*'? 

210° 

-  .86 

420° 

.50 

-6.1 

-.50 

.86 

-9.3 

240° 

-.50 

480° 

-.50 

-7.5 

-.86 

.86 

-12.9 

270° 

0 

540° 

-1 

-5 

-1 

0 

-10 

300° 

.50 

600° 

-.50 

2.5 

-.86 

-.86 

^'^ 

330° 

.86 

660° 

.50 

11.1 

-.50 

-  .86  * 

^'0.7 

360° 

1 

720° 

1 

15 

0 

0 

0 

To  obtain  the  rectangular  equation  from  the  parametric  equa- 
tions, the  parameter  must  be  eliminated.  The  method  used 
depends  upon  the  example. 


PARAMETRIC  EQUATIONS  AND  LOCI  207 

EXAMPLES 

1.  Find  the  rectangular  equation  of  the  curve  whose  parametric  equa- 
tions are 

(4)  x  =  2t  +  3,     y  =  ir--4. 

Solution.  The  first  equation  may  be  solved  readily  for  t.  We  find 
f  =  |(x  —  3),  andsubstitutingin  the  second  equation  gives  J/ =  i(x  —  3)^—4; 
or,  expanding  and  simplifying,  a;^  —  6x  —  8?/  —  23  =  0,  a  isai'abola. 

2.  Find  the  rectangular  equation  of  the  curve  whose  parametric  equa- 
tions are 

(5)  X  =:  3  -h  4  cos  ^,     ?/  =  3  sin  0. 

Solution.  Remembering  that  sin^  ^+  cos-  ^=:  1,  we  solve  the  first  equa- 
tion for  cos^,  the  second  for  sin  0.    This  gives 

(6)  cos^  =  ^(x-3),     sin0  =  ^y. 
Hence  the  rectangular  equation  is 

^''  16       ^9        ' 

an  ellipse. 

PROBLEMS 

1.  Plot  the  following  parametric  equations,  t  and  0  being  variable 
parameters.   Find  the  rectangular  equation  in  each  case  : 
"^  (a)  X  =  i  -  1,  2/  =  4  —  (2.  ( i )  X  =  cos  (9,  2/  =  cos  2  ^. 

(b)  X  =  2  ^2  _  2,  7/  =  i  -  3.  ( j  )  X  =  1  sin  0,  y  =  sin  2  0. 

(c)  X  =  3  cos  0,  y  =  sin  0.  ( k )  x  =  1  —  cos  0,  y  =  \  sin  I  0. 

(d)  X  =  3  tan  0,  y  =  sec0.  {\  )  x  =  St^,  y  =  St  -  t^. 

■    (e)  x  =  2t  v  =  --  (m)  x  =  2sini9  +  3cos(9,  ^/^siiK?. 

^'  t'  (n)  x  =  2cosi9  +  l,|/=sin^  +  4cosi9. 

(f )  X  =  2  +  sin  0,y  =^2cos0.  (o)  x  =  t-  t'\  y  =  t  +  t^. 

{g)x  =  lt^,y  =  lt.  (p)x  =  3-2i,z/  =  l  +  l 

(h)    X=    «2_2i,  ,y  =  l_<2.  ^^   '  '   ^  ^   t 

\l    2.  Plot  the  following  parametric  equations  : 

(a)  X  =  2  r  cos  0  —  r  cos  2  0,  ^  y  =  2r  sin  0  —  r  sin  2  0. 

(b)  X  =  3  r  cos  0  +  r  cos  S0,  y  =  3rs\n0  —  r  sin  3  0. 

(c)  X  =  3  r  cos  0  —  r  cos  S0,  y  =  Sr  sin  ^  —  r  sin  3  0. 

(d)  X  —  r  cos 0  —  r  cos  2  0,  y  =  rsin0  —  r  sin 2 0. 

(e)  X  =  2  r  cos  0  +  ^r  cos  2  0,  y  =  2r  sin  0  —  \r  sin  2  6. 


208 


NEW  ANALYTIC  GEOMETRY 


(f) 


X  =  a{6  ■ 


sm  6'), 
cos^). 


.  ,    (x-=  a{Q  -^  smO), 
^^'  \y  =  a{\-cosd). 


Y 

^  a 

.^c 

a 

\ 

A 

\/ 

w^ 

^^ 

a 

\/ 

r^^ 

V 

0 

A' 

X 


D,    CUSP   AT    ORIGIN 

CYCLOID,    VERTEX   AT    ORIGIN 

(h)  x  =  aO  —  ^asin^, 

y  =  a  —  \a  cos  6. 

{ i  )  x  =  ae  -2a  sin 6, 

y  —  a  —  'ia  cos  6. 

( j )  X  =  r  cos  9  ■\-  r  0  sin  6^  y  =  r  sin  6  —  rO  cos  ^. 
( k )  X  =  4  r  COS  ^  —  r  cos  4  ^,  y  —  ir  sin  ^  —  r  sin  4  ff. 

(1)  X  =  alogi,  ?/  =  |a/i  +  -|. 

(ni)  X  =  f  +  sini,  y  =  1  +  cos^. 

{ n )  X  =  2  cos  t  +  t,  y  =  3  cos  i  +  sin  2  ^. 

(  o )  X  =  b  cos'^  0^  y  =  a  tan  ^. 

V  81.  Various  parametric  equations  for  the  same  curve.    When  the 
rectangular  equation  of  a  curve  is  given,  any  number  of  para- 
metric equations  may  be  obtained  for  the  curve. 
For  example,  given  the  ellipse 

(1)  4.T^  +  v/-^  =  16. 

Let  a;  =  2  cos  6,  where  ^  is  a  variable  parameter.    Substitut- 
ing in  (1), 

16  cos^^  +  y- =  16,    or   2/^=16(l-cos"'^)=16sin-^. 
Hence  the  equations 

(2)  a;  =  2cos^,     ?/  =  4sin^, 

are  parametric  equations  of  the  ellipse  (1). 

Again,  substitute  in  (1), 

y  =  fx  +  4, 

where  t  is  a.  variable  parameter. 
This  gives 

(3)  4  a;2 -K  !^V -f  8  At  +  16  =  16,     or  (4 -f- i;')^-'-!- 8  ^a;  =  0. 

(4)  ■'^  =  -TTf' 


PARAMETRIC   EQUATIONS  AXD  LOCI 


209 


Substituting  this  value  in  y  =  fee  +  4  and  reducing, 


(5) 


y 


16  -  4  j'^ 

4.  +  f' 


Hence  the  equations  (4)  and  (5)  are  also  parametric  equa^- 
tions  of  the  ellipse. 

The  point  is  :  We  obtain  parametric  equations  by  setting  one 
of  the  coordinates  equal  to  a  function  of  a  'parameter,  substitut- 
ing in  the  given  rectangular  equation  and  solcing  for  the  other 
coordinate  in  terms  of  the  parameter. 

To  obtain  simple  parametric  equations  we  must,  of  course, 
assume  the  right  function  for  one  coordinate.  No  general  rule 
applicable  to  all  cases  can  be  given  for  this  purpose,  but  the 
study  of  the  problems  below  will  aid  the  student. 

Many  rectangular  equations  difficult  to  plot  are  treated  by 
deriving  parametric  equations  and  plotting  the  latter. 


EXAMPLES 

'    1.  Draw  the  locus  of  the  equation 

(6)  x3  +  ?/3  _  3  ctxy  =  0. 

Solution.  Set  y  =  te,  where  I  is  the  parameter.   Then,  from  (6), 

(7)  a:3  +  <3j.3  _  3  a^a;2  =  0. 

Dividina:  out  the  x^,  solving  for  x, 
and  remembering  that  y  =  te,  we  obtain 
the  desired  parametric  equations 

3  at  3  ai^ 


(^) 


1  +  i^ 


y  = 


1  + 


The  locus  is  the  curve  of  the  figure, 
called  the  folium  of  Descartes. 

The  line  drawn  in  the  figure  is  an 
oblique  asymptote.  Its  equation  is 
X  4-  y  +  a  =  0. 

The  parameter  t  in  (7)  is  obviously  the  slope  of  the  line  y 
is,  of  the  line  joining  a  point  on  the  curve  and  the  origin. 


tx ;  that 


A 


210 


NEW  ANALYTIC   GEOMETRY 


The  reason  for  assuming  the  relation  y  =  tx  in  the  preceding 
example  is  that  x'-  divides  out  in  (7),  leaving  an  equation  of  the 
first  degree  to  solve  for  x.  Problems  1  (a),  (d),  (e),  (f),and  (j) 
below  are  worked  on  the  same  principle.  In  many  cases  trigo- 
nometric functions  are  employed  with  advantage,  as  in  (b) 
and  (c). 

PROBLEMS 

1.  Find  parametric  equations  for  each  of  the  following  curves  by 
making  the  substitution  indicated  in  the  given  equation.    The  parameter 
is  t  or  0,  as  the  case  may  be.   Plot  the  locus. 
\\  (a)  2/2  =  4 a;2  _  x3,  ?/  =  tx.  Ans.   x  =  4:  —  t^,y=:4:t  —  t^. 

(b)  x^y^  =  b-x'^  +  a^  y'^,  x  =  a  sec  6.  Ans.   y  =  b  esc  d. 

I  (c)  x^y^  =  d-y-  —  b-x-,  x  —  asm 6.  Ans.   y  =  6tan^. 

y  (d)  y^  -2  ax-—  x^,  y  —  tx.  ^ 


0     "  X 

(h)  x^  +  y^  =  a^,x  —  a co^O. 

PARABOLA 


\    (e)  2/2  (2  a  —  x)  =  x^,  y  —  tx. 


CISSOID    OF    DIOCLES 

(f)  2/2  =  x2|+-5,2/  =  te. 

2  —  X 

2  <2  _  2            2  i3  _  2  « 
Ans.  X  — 1  y  = • 

(g)  x2  +  xy  +  22/2  +  2x  +  1=0, 

X  =■  tV  —  1.  '  i22 

,                        2  +  ?-                      1         -  ( i  )  x*+  yi  =  a^,x  =  a  sin^  6. 
Ans.  X  =  — ,  y  =  — •  ^    ' 

t1  J^  t  +  2  t-  -{■  t  -ir  2  HYPOCYCLOID   OF    FOUR   CUSPS„ 


PARAMETRIC  EQUATIONS  AND  LOCI 


211 


( j )  a;*  +  2  ax^y  —  ay^  =  0 ,  y  =  tx. 

(k)  (a;2  +  y2  +  4ay-  a-)  {x^  -  a^)  +  4  a"y^  =  0,  x^  =  a^  -  i-y-. 

(1)  x''  =  y{y-2f,y-2  =  tx. 

(in)     ( X2  -  I  62)2  ^  y2  (a.2  _  ^2)  =  0,  X2  =    1  ^2  ^.  ty. 

82.  Locus  problems  solved  by  parametric  equations.  Parametric 
equations  are  important  because  it  is  sometimes  easy  in  locus 
problems  to  express  the  coordinates  of  a  point  on  the  locus  in 
terms  of  a  parameter,  when  it  is  otherwise  difficult  to  obtain 
the  equation  of  the  locus.  The  following  examples  illustrate 
this  statement : 

EXAMPLES 

1.  ABP  is  a  rigid  line.  The  points  A  and  B  move  along  two  perpen- 
dicular intersecting  lines.   What  is  the  locus  of  the  point  P  on  ^iJ  ? 

In  the  figure,  A  moves  on  XX',  B  moves  on 
YY' ;  required  the  locus  of  the  point  P(x,  y). 

Solution.  Take  the  coordinate  axes  as  indi- 
cated, and  consider  the  line  in  any  one  of  its 
positions.  Choose  for  parameter  the  angle 
XAB  =  e. 

Let  AP  -a,     PB  =  b. 

Now  OM  =  X,     MP  =  y. 

In  the  right  triangle  MPA, 

.    n       MP      y 

sni  6  = =  -  • 

PA      a 

In  the  right  triangle  BSP,  Z.  PBS  =  6. 

.-.  cos  PBS  =  cos  0  — =  -  • 

BP      b 
From  (1)  and  (2), 

(3)  X  =  b  cos  0,     y  =  a  sin  0. 

These  are  the  parametric  equations  of  the  locus. 

Squaring  (1)  and  (2)  and  adding, 


(1) 


(2) 


62  '^  a^ 


=  1. 


Hence  the  point  P  moves  on  an  ellipse  whose 
axes  2  a  and  2  b  lie  along  the  given  perpendicular  lines. 

A  method  commonly  employed  for  drawing  ellipses  depends  upon  this 
result.   The  instrument  consists  of  two  grooved  perpendicular  bars  X'X 


212 


NEW  ANALYTIC   GEOMETRY 


and  YY'  and  a  crossbar  ABP.  At  A  and  B  are  screw  nuts  fitting  the 
grooves  and  adjustable  along  ABP.  If  the  crossbar  is  moved,  a  pencil 
at  P  will  describe  an  ellipse  whose  semiaxes  are  PA  and  PB. 

©  2.  The  cycloid.    Find  the  parametric  equations  of  the  locus  of  a  point 
P  on  a  circle  which  rolls  along  the  axis  of  x. 


Solution.  Take  for  origin  a  point  0  at  which  the  moving  point  P 
touched  the  axis  of  x.  Let  the  circle  drawn  be  any  position  of  the  rolling 
circle.  Let  a  be  the  radius  of  the  circle  and  take  for  the  variable  param- 
eter 6  the  variable  angle  CBP.   Then 

PC  =  a  sin  0,     CB  =  a  cos  0. 

By  definition,  OA  —  arc  AP  =  ad. 

[For  an  arc  of  a  circle  equals  its  radius  times  the  subtended  angle, 
from  the  definition  of  a  radian.] 

Hence  from  the  figure,  if  (x,  y)  are  the  coordinates  of  P, 
X  =  OD  =  OA  -  PC  =  ad  -  a  sin  6,  y  =  BP  =  AB  -  CB  =  a  -  a  cos  (9. 


(4) 


(X 


a{9  —  sin  6), 
\^y  =  a{l  —  cosO). 

These  are  the  parametric  equations  of  the  cycloid. 

The  cycloid  extends  indefinitely  to  the  right  and  left  and  consists  of 
arcs  equal  to  OMN.  3/ 

Construction  of  the  cy- 
cloid. The  definition  of  the 
cycloid  suggests  the  follow- 
ing simple  construction : 

Lay  off  0N=2Tra  ^  cir- 
cumference of  the  generat- 
ing circle.    Draw  the  latter  touching  at  C,  the  middle  point  of  ON. 
Divide  OC  into  any  number  of  equal  parts,  and  the  semicircle  C3f  into 


PARAMETRIC  EQUATIONS  AND  LOCI         "    213 

the  same  number  of  equal  arcs.    Letter  as  in  the  figure.    Through  M\, 
M„,  etc.,  draw  lines  parallel  to  ON.    Lay  off 

M^Dj^  =  CC^,     M^D^  =  CCg,     M^D^  =  CC^,  etc. 

Then  D^,  D„,  Dg,  etc.  are  points  on  the  cycloid. 

For,  let  the  generating  circle  roll  to  the  left,  the  point  M  tracing  the 
curve.  When  the  circle  touches  ON  at  C^,  M  will  lie  on  a  level  with  jVj, 
and  at  a  distance  to  the  left  of  ilf^  equal  to  CC-^^.   Similarly  for  D^.,  Dg,  etc. 

The  arc  MN  of  the  cycloid  may  be  constructed  by  using  CM  as  an 
axis  of  symmetry. 


n\  3.  The  hypocycloid  of  four  cusps.  Find  the  parametric  equations  of 
the  locus  of  a  point  P  on  a  circle  which  rolls  on  the  inside  of  a  fixed 
circle  of  four  times  the  radius. 

Y 


Solution.    Take  the  center  of  the  fixed  circle  for  the  origin  and  let  the 
X-axis  pass  through  a  point  A  where  the  tracing  point  P  touched  the 

large  circle.    Then  OA  =  4  C2>,  by  hypothesis. 


_,„      OA      a     ^ 

CB  = =  -  .    Draw 

4        4 


the  rolling  circle  in  any  of  its  positions.   Take  for  the  variable  parameter  0 
the  Z  A  OB.   Then  Z  BCP  =  4  ^. 


214  NEW  ANALYTIC   GEOMETRY 

[For,  by  hypothesis,  arc  PB  —  arc  A  B  ;  and,  from  the  definition  of  a 
radian,  arc  P5  =  - Z-BCP,  arc ^5= a^.    .-.  -  Z  BCP=aO,orZBCP=4d.] 
But  ZOCE+  ZECP  +  ZPCB  =  -jr. 

.:  --0  +  ZECP  +  4e  =  7r. 

2 

Whence  Z  ECP  =  --S0. 

Now  OF  =  x,  FP  =:  y. 

From  the  figure, 


(5) 


0F=  OE  +  DP, 
FP  =  EC  -  CD. 

Finding  the  lengths  of  the  segments  in  the  right-hand  members, 

OE  =  OCcos0  =  —  cos  0,  EC  =  OC sin  0  =  —  sin  ff. 
4  4 

DC=  CP  cos  (^^-3(9')  =  - sin  3  (9,    .  (by  31,  p.  3) 

J)P=  CP  sin/- -3  (9")  = -cos  3^.  (by  31,  p.  3) 

Substituting  in  (5), 

f  X  =  f  a  cos 0  +  ia cos 3 d, 

ly  =  iashi  0  —  ianinS 6. 
These  are  parametric  equations  for  the  hypocycloid  of  four  cusps. 
Anotlier  form  of  (6)  from  wliich  tlie  rectangular  equation  may  easily 
be  derived  is  obtained  by  expressing  cos  3  0  and  sin  3  0  in  terms  of  cos  0 
and  sin  0  respectively.   Thus, 

cos  3  (9  =  cos  (2  ^  +  ^)  =  cos  2  (9  cos  ^  -  sin  2  (9  sin  0      (by  35,  p.  3) 
=  (2  cos2  ^  —  1)  cos  ^  —  2  sin2  0  cos  0 
=  2  cos3 ^  _  cos^  -  2  (1  -  cos'^ 0)  cos  0 
=  4  cos^^—  3  cos^. 
sin  30  =  sin {2 0  +  0)  =  sin  2  6'  cos ^  +  cos 2  ^  sin 0      (by  33,  p.  3) 
=  2  sin  0  cos2(9  +  (1  —  2  sin2(9)  sin  0 
=  2  sin  (9  (1  -  sin^  ^)  +  sin  ^  -  2  sin3  0 
=  3sin(9-4sin3^. 

Substituting  in  (6)  and  reducing,  the  result  is 

(7)  X  =  aGOs^0,     y  =  asm^0. 
From  these,  x^  =  aJ  cos^^,     ?/3"  —  a^  sin-^.   Adding, 

(8)  x^  +  y^  =  a*, 

which  is  the  rectangular  equation  of  the  hypocycloid  of  four  CMsps 


PARAMETRIC  EQUATIONS  AND  LOCI 


215 


PROBLEMS 

In  the  following  problems  express  x  and  y  in  terms  of  the  parameter 
and  the  lengths  of  the  given  lines  of  the  figure.    Sketch  the  locus. 

1.  Find  the  parametric  equations  of  the  ellipse,  using  as  parameter 
the  eccentric  angle  0,  that  is,  the  angle  between  the  major  axis  and  the 
radius  of  the  point  B  on  the  major  auxiliary  circle  (p.  164)  which  has 
the  same  abscissa  as  the  point  P  (x,  y)  on  the  ellipse.    (See  figure.) 

Ans.   X  =  a  cos  ^,y  =  b  sin  ^. 


Y 

"^^""^--.^ 

1  1     \± 

11              ^ 

;mm^x 


2,  In  the  figure,  ABP  is  a  rigid  equilateral  triangle.    A  moves  on 
YY',  B  moves  on  XX'.   Find  the  locus  of  the  vertex  P. 

Ans.   x  =  acos(9+  a  cos  (120°- ^),  ?/ =:  a  sin(120°- (?). 
Ellipse,  x^  —  V3  xy  ■\-y'^—\a^. 

3.  Two  vertices^  and  B  of  a  rigid  isosceles  right  triangle  ABP  move 
on  perpendicular  lines.    Find  the  locus  of  the  vertex  P. 

Ans.   x  =  acosd+  a  sin  0,  y  =  a  cos  0.     Ellipse,  X'—2xy  +  2y^=  a^. 


i.  AB  is  a,  fixed  line  and  R  a  fixed  point.  Draw  RQ  to  any  point  Q 
in  AB  and  erect  the  perpendicular  QP,  making  QP  -i-  QR  equal  to  a  con- 
stant e.   What  is  the  locus  of  P  ?  ^2       ^,2 

Ans.   x-pcotO,y  =  epcscff.     Hyperbola,^ -^  =  — 1. 


216 


NEW  ANALYTIC   GEOMETRY 


5.  AB  is  a  fixed  line  and  0  a  fixed  point.  Througli  0  draw  OX 
parallel  to  AB  and  ON  perpendicular  to  AB.  Draw  a  line  from  0 
through  any  point  Q  in  AB.  MSrk  on  this  line  a  point  P  such  that 
MP  -  NQ,  MP  being  J.  to  OX.    What  is  the  locus  of  P? 

Ans.   X  —  a  cot^^,  y  =  a  cot^.   Parabola,  y^  =  ax. 

li?(a,6) 


Ans.    'i, ' 


6.  Through  the  fixed   point  U  (a,  6)  lines   are  drawn   meeting  the 
coordinate  axes  in  A  and  B.  What  is  the  locus  of  the  middle  point  ot  AB? 

Ans.    2x  =  a ,  2y  =  h  —  at,  where  t  =  slope  of  AB. 

Equilateral  hyperbola,  (2  x 

7.  Find  the  locus  of  a  point  Q  on 
the  radius  BP  (Fig.,  Ex.  2,  p.  212) 
itBQ  =  b. 

fx  =  aO  —  b  sin  9, 

[^y  =  a  —  b  cos^. 
The  locus  is  called  a  prolate  or  cur- 
tate cycloid  according  as  fl|y^  greater 
or  less  than  bk 

Describe  a  •construction  for  the 
curve  analogous  to  that  given  for 
the  cycloid  in  Art.  82. 

8.  Given  a  string  wrapped  around 
a  circle ;  find  the  locus  of  the  end 
of  the  string  as  it  is  unwound. 

Hint.  Take  the  center  of  the  cir- 
cle for  origin  and  let  the  .T-axis  pass 
through  the  point  A  at  which  tlie  end 
of  the  string  rests.  If  the  string  is  im- 
wound  to  a  point  B,  let  ZAOIJ:=d. 

(See  figure.)  _.,.,.,      fx  =  rcos^ -t- r^sio^, 

Ans.   The  involute  of  a  circle   -{  .    ^       „        ^ 

(_ij  =  r  sin  ff  —  rff  cos  0, 


o 


PARAMETRIC  EQUATIONS  AND  LOCI  217 

9.  A  circle  of  radius  r  rolls  on  the  inside  of. a  circle  whose  radius 


is  r'.    Find  the  locus  of  a  point  on  the  rolling  circle. 
Ans.   The  hypocycloid  "V  '^'^'^^M  ^• 

r  Y 


('•' 


r)  cos  6  -\-  r  cos 6, 

r 

t'  —  T 

y  —  {r'  —  r)  sin  6  —  r  sin 6. 

The  curve  is  closed  when  r  and  r'  are 
commensurable.  The  hypocycloid  of 
four  cusps,  p.  213,  is  a  special  case. 

Describe  a  construction  for  the  curve 
analogous  to  that  given  for  the  cycloid 
in  Art.  82. 

rr>    10.  A  circle  of  radius  r  rolls  on  the  outside  of  a  circle  whose  radius- 
is  r'.    Find  the  locus  of  a  point  on  the 
rolling  circle. 

Ans.   The  epicycloid 

r'  4-  r 


Kc  =  (r'  +  r)  cos  9  —  r  cos 
I  y  =  (r'  +  r)  sin  ^  —  r  sin 


r 

•/  +  r 


e. 


The  curve  is  closed  when  r  and  r'  are 
commensurable. 

Describe  a  construction  for  the  curve 
analogous  to  that  given  for  the  cycloid 
in  Art.  82. 

11.  Given  a  fixed  point  0  on  a  fixed  circle  and  a  .fixied  line  AB,  Draw 
the  X-axis  through  0  perpendicular  to  AB 
and  the  ?/-axis  through  O  parallel  to  AB. 
Draw  any  line  through  0  to  meet  AB  m 
L  and  the  fixed  circle  in  S.  Draw  LP  II 
to  OX  to  meet  SM  drawn  II  to  OY.  Re- 
quired the  locus  of  P. 

Ans.   X  =  6  cos^ ^,  ?/ =  a  tan  ^. 

Cubic,  xy^+  a^x  —  a%  =  0.' 

Give  a  full  discussion  of  the  equation. 
Show  that  the  y-axis  is  an  asymptote. 
What  modifications,  if  any,  are  necessary 
in  the  equations  when  yl  J5  is  a  tangent  ? 
when  AB  does  not  intersect  the  circle  ? 


218 


NEW  ANALYTIC   GEOMETRY 


J 


12.  OB  is  the  crank  of  an  engine  and  AB  the  connecting  rod.    B  moves 
on  the  crank  circle  whose  center  is 
0,  and  A  moves  on  the  fixed  line 
OX.  What  is  the  locus  of  any  point 
P  on  AB  ? 

Ans.   x  =  bcos0 
+  Vr2  —  (a  +  Oy^  sin'-^6',  y  =  a  sin  0. 

Ellipse,  when  r  =  a  +  b;  other- 
wise an  egg-shaped  curve. 

13.  OB  is  an  engine  crank  re- 
volving about  O,  and  A  Bis  the  con- 
necting rod,  the  point  A  moving  on   OX. 
Draw  AP  ±  to  OX  to  meet  OB  produced 
at  P.*   What  is  the  locus  of  P  ? 

Ans.   X  =  r  cos  ^-|-  Vc'-^  —  r'-'sin'"^, 

y  =  r  sin  0  +  tan  OVc-  —  r^  sin"-^ 
When  c  =  r,  the  locus  is  the  circle 

a;2  -f  7/2-  4^2 

83.  Loci  derived  by  a  construction 
from  a  given  curve.  Many  important 
loci  are  defined  as  the  locus  of  a  point 
obtained  by  a  given  construction  from  a  given  curve.  The 
method  of  treatment  of  such  loci  is  illustrated  in  the  follow- 
ing examples. 

EXAMPLES 

1.  Find  the  locus  of  the  middle 
points  of  the  chords  of  the  circle 
X-  +  y"  =  25  which  pass  through 
P2(3,4). 

Solution.   Let  P^(Xi,  ?/j)  be  any 
point  on  the  circle. 
(1)  •••  x2  -1-  2/f  =  25. 

Then  a  point  P  (x,  y)  on  the  locus 
is  obtained  by  bisecting  PjP,.  By 
(IV),  Art.  13, 

.-.  Xi  =  2a:-3,       v^^2y-4. 
*  P  \s  the  "  instantaneous  center  "  of  the  motion  of  the  conuet;ting  rod. 


rA 

k 

^ 

•^ 

;^ 

^ 

^!-p,J 

; 

/ 

/ 

/ 

/ 

' 

y 

\ 

/ 

1 

/ 

V 

'h 

•y, 

A' 

(; 

— 

^ 

X 

y 

1 

\ 

xr, 

n) 

\ 

^  I 

^ 

^ 

^ 

^ 

~1 

— 

— 

— 

y 

— 

— 

— 

— 

— 

, 

— 

PARAMETRIC  EQUATIONS  AND  LOCI  219 

Substituting  in  (1). 

■  (2x- 3)2 +  (22/ -4)2  =  25, 

or  x^  +  y^  —  S X  —  4  y  =^  0.    Ayis. 

The  locus  is  a  circle  constructed  upon  OP^  as  a  diameter. 

2.  The  witch.  Find  the  equation  of  the  locus  of  a  point  P  constructed 
as  follows  :  Let  OA  be  a  diameter  of  the  circle  X'  +  ij'^  —  2  ay  =  0,  and 
let  any  line  OB  be  drawn  through  0  to  meet  the  circle  at  P^  anil  the 
tangent  at  A  at  B.  Draw  PjP  ±  to  OA  and  BP  II  to  OA.  Required 
tlie  locus  of  P. 

Solution.    Let  (x,  y)  be  the  coordinates  of  P  and  (Xj,  y^)  of  Pj. 
Then  the  coordinates  of  P^  (a;^,  ?/^)  must  satisfy  the  equation 

a;2  j^  yi  —  2ay  —  0. 

(2)  .-.  x2  + 2/2- 2 02/1  =  0. 
From  the  figure, 

(3)  y,  =  y. 

From   the    similar    triangles 
OCP^  and  0MB  we  have 

OC       GP,  X, 

= -1    or    — 

OM      MB  X 

[For  OC  =  Xj,     OJVf : 


A 

B 

yf^ 

">> 

A 

yT " 

^^ 

--  1 

^[ 

V 

"^nlT" 

--^K' 

h 

(4) 


2a 


CP^  =  y^,     MB  =  2a.li 
Solving  (3)  and  (4)  for  Xj  and  t/j,  we  obtain 

(5)  "^^'S'     ^»"^" 

Substituting  from  (5)  in  (2), 

2ay  =  0, 


^    y  n 

4a^ 


or 
(6) 


?/(x2  +  4a2)  =  8a3. 


The  locus  of  this  equation  is  known  as  the  witch  of  Agnesi. 
The  method  followed  in  Examples  1  and  2  may  evidently  be 
described  as  follows  : 

Rule  fo7'  finding  the  equation  of  a  locus  derived  by  a  construc- 
tion from,  a  given  curve. 


220 


NEW  ANALYTIC  GEOMETRY 


First  step.  The  construction  will  give  rise  to  a  figure  from 
which  we  viay  find  expressions  for  the  coordinates  of  any  point 
P^(x^,  y^  on  the  given  curve  in  terms  of  a  point  P(x,  y)  on  the 
required  curve. 

Second  step.  Substitute  the  results  of  the  first  step  for  the  coor- 
dinates x^  and  y^  in  the  equation  of  the  given  curve  and  sivipdify. 
The.  result  is  the  required  equation. 


PROBLEMS 

1.  Find  tlie  locus  of  a  point  whose  ordinate  is  half  the  ordinate  of  a 
point  on  the  circle  x^  +  y- —  64.  Arts.   The  ellipse  x^  +  4  2/2  =  64. 

2.  Find  the  locus  of  a  point  which  cuts  off  a  part  of  an  ordinate  of 
the  circle  x^  -\-  y-  =  a^  whose  ratio  to  the  whole  ordinate  is  b  :  a. 

Ans.   The  ellipse  b^x'^  +  a^y"^  =  a^b^. 

3.  Find  the  locus  of  the  middle  points  of  the  chords  of  (a)  an  ellipse, 
(b)  a  parabola,  (c)  a  hyperbola  which  pass  through  a  fixed  point  P^i^zi  ^2) 
on  the  curve. 

Ans.  A  conic  of  the  same  type  for  which  the 
values  of  a  and  6  or  of  p  are  half  the  values  of  those 
constants  for  the  given  conic. 

4.  Lines  are  drawn  from  the  point  (0,  4)  to  the 
hyperbola x'^  —4y^  —  16.  Find  the  locus  of  the  points 
which  divide  the.se  lines  in  the  ratio  1:2. 

Ans.    3x2-12?/2  + 64  2/-90f  =  0. 

5.  A  chord  OP^  of  the  circle  x^  +  y^-2ax  =  0 
meets  the  line  x  =  2a  at  a  point  A.  Find  the  locus 
of  a  point  P  on  the  line  OP^  such  that  OP  =  P^A. 

Ans.  The  cissoid  of  Diodes  y'^{2a  —  x)  =  x^  (see 
figure). 

6.  DIX  is  the  directrix  and  F  the  focus  of  a  given  conic  (Art.  72). 

Q  is  any  point  on  the  conic.  Through  Q  draw  QN  ±  to  the  axis  of  the 
conic  and  construct  P  on  NQ  so  that  NP  =  FQ.    What  is  the  locus  of  P  ? 

Ans.    A  straight  line. 

84.  Loci  using  polar  coordinates.  When  the  required  locus  is 
described  by  the  end-point  of  a  line  of  variable  length  whose 
other  extremity  is  fixed,  polar  coordinates  may  be  employed  to 
advantage. 


PARAMETRIC  EQUATIONS  AND  LOCI 


221 


EXAMPLE 

The  conchoid.  Find  the  locus  of  a  point  P  constructed  as  follows  : 
Through  a  fixed  point  0,  a  line  is  drawn  cutting  a  fixed  line  ^J5  at 
Pj.  On  this  line  a  point  P  is  taken  so 
that  P^P  —  ±^,  where  6  is  a  constant. 

Solution.  The  required  locus  is  the 
locus  of  the  end-point  P  of  the  line  OP, 
and  0  is  fixed.  Hence  we  use  polar  coor- 
dinates, taking  0  for  the  pole  and  the 
perpendicular  OM  to  AB  for  the  polar 
axis.    Then 

(1)  OP  =  p,     ZMOP  =  d. 
By  construction, 

(2)  p=  0P=  0P^±  b. 
But  in  the  right  triangle  OMP^, 

(3)  OP^  =  OM  sec  ZMOP^  =  a  sec  0. 
Substituting  from  (3)  in  (2), 

(4)  p  =  a  sec  0  ±b. 
The  locus  of  this  equation  is  called  the  conchoid  of  Nicomedes.   It  has 

three  distinct  forms  according  as  a  is  greater,  equal  to,  or  less  than  6. 


PROBLEMS 

1.  OA  is  a  diameter  of  a  fixed  circle,  and  OB  is  any  chord  drawn  from 
the  fixed  point  0.    In  the  figure  below,  BP  =  AB.    Find  the  locus  of  P. 

Ans.   The  circle  p  =  a  (sin  0  +  cos^). 


2.  The  chord  OB  of  a  fixed  circle  drawn  from  0  is  produced  to  P, 
making  BP  =  diameter  =  a.  What  is  the  locus  of  P  ? 

Ans.   The  cardioid /o  =  «(1 -I- cos^).. 


222 


NEW  ANALYTIC  GEOMETRY 


3.  In  problem  2,  if  BP  =  any  length  =  b,  the  locus  of  P  is  the  limapon 
of  Pascal,  p  =  b+  a  cos^.  The  limagon  has  three  distinct  forms  accord- 
ing as  b  =  a.  In  the  figure  on  p.  124,  b  <  a.  The  rectangular  equation 
is  (x2  +  2/2  +  aa;)2  =  62  (3.2  +  ^z^). 

4.  F  is  the  focus  and  DD'  the  directrix  of  a  conic  (figure  below).  Q  is 
any  point  on  the  conic.   On  the  focal  radius  FQ  lay  off  FP  =  QM,  where 


QM  is  II  to  DD'.   Find  the  locus  of  P  (see  Art.  72). 


Ans.  p  = 


ep  sin  ff 
1  —  e  cos  ( 


^■"^^^'Cb.^ 

L 
C 

0 

^^^zKi'y^ 

X 
M 

5.  Lines  are  drawn  from  the  fixed  point  0  on  a  fixed  circle  to  meet  a 
fixed  line  LM  which  is  ±  to  the  diameter  through  O.  On  any  such  line  OG 
lay  off  OP  =  BC.   What  is  the  locus  of  P?        Ans.  p  =  bsecO  —  acos0. 

Draw  the  locus  for 
b>a,b<a,  and  b  =  a. 
In  the  last  case  the 
curve  is  the  cissoid 
(Problem  5,  p.  220). 

6.  0  is  the  center 
of  a  fixed  circle  and  A 
a  fixed  interior  point. 
Draw  any  radius  OB, 
connect^  and  ii, and 
draw^P±  to  AB  to 
meet  OB  at  P.  Re- 
quired the  locus  of  P. 

.  e  — acos^ 

A  n.s.  p  =  e . 

c  cos  ff  —  a 
if  OB  =  a,OA=ze. 
Draw  the  locus. 

7.  A  line  is  drawn  from  a  fixed  point  O  meeting  a  fixed  line  in  Pj.  Find 
the  locus  of  a  point  P  on  this  line  such  that  OP^  •  OP  =  a^.     Ans.    A  circle. 


PARAMETRIC  EQUATIONS  AXD  LOCI 


223 


8.  A  line  is  drawn  through  a  fixed  point  O,  meeting  a  fixed  circle  in 
Pj  and  P„.    Find  the  locus  of  a  point  P  on  this  line  such  that 

OP  =  2  OP^  ■  OP^  -^{OPi  +  OP^).     Ans.  A  straight  line. 

9.  In  I'^.x.  1,  Art.  82,  find  the  locus  of  the  foot  of  the  perpendicular  from 
the  origin  upon  ^5.   Ans.  Four-leaved  rose  p  =  ^  (a  —  6) sin  2  ^  (see  figure). 

r 


FOUR-LEAVED    ROSE 


PARABOLIC    Sl'IUAl, 


10.  Let  the  x-axis  cut  the  circle  x^  +  y'^  =  a-&tA.  An  arc  AB  is  laid 
off  on  the  circle  equal  to  the  abscissa  Xg  of  a  point  (Xq,  y^)  on  the  parabola 
y^  =  4  ex,  and  the  radius  OB  is  produced  to  P  making  BP  =  y^.  Show 
that  the  locus  of  P  is  the  parabolic  spiral  {p  —  a)-  =  4ac0  (see  figure). 

y  85.  Loci  defined  by  the  points  of  intersection  of  systems  of  lines. 
If  two  systems  of  lines  involve  the  same  parameter,  the  lines 
belonging  to  the  same  value  of  the  parameter  are  called 
corresponding  lines.  Many  loci  are  defined,  or  may  be  easily 
regarded  as  the  locus  of  the  points  of  intersection  of  such 
lines.    The  method  of  treatment  will  now  be  illustrated. 


EXAMPLES 

1.  Find  the  locus  of  the  foot  of  the  perpendicular  drawn  from  the  ver- 
tex of  a  parabola  to  the  tangent.    (See  the  figure  on  p.  224). 

Solution.  Taking  the  typical  equation  y^  =  2px,  the  equation  of  a 
tangent  AB  in  terms  of  the  slope  t  is  (Art.  78) 

(1) 


y 


2t 


The  equation  of  the  perpendicular  OP  is 

(2)  y 


1 

-X. 

t 


224 


NEW  ANALYTIC   GEOMETRY 


Equations  (1)  and  (2)  define  the  two  systems  of  lines  in  the  parameter  t. 

The  locus  of  the  point  of  intersection  P  of  corresponding  lines  is  required. 

Solving  (1)  and  (2)  for  x  and  y, 

(3)  X= P,y  =  -1—. 

These  are  the  parametric  equations  of 
the  required  locus. 

The  rectangular  equation  is  found 
thus  : 

X 

From  (2),  t  = Substituting  in 

y 

the  first  equation  of  (3)  and  reducing, 
2/2  (x+  ip)--x^. 
Comparison  with  the  answer  to 
Problem  5,  p.  220,  shows  that  the 
locus  is  a  cissoid. 

The  method  of  solving  Example  1  may  be  summed  up  in  the 

Rule  to  find  the  erjiiaflon  of  the  locus  of  the  points  of  intersec- 
tion (f  correspond incj  lines  of  two  systems. 

First  step.  Find  the  equations  of  the  two  systems  of  lines 
defining  the  locus  in  terms  of  the  same  parameter. 

Second  step.  Solve  these  equations  for  x  and  y  in  terms  of  the 
pjara meter.    This  gives  the  parametric  equations  of  the  locus. 

If  only  the  equation  in  rectangular  coordinates  is  required,  it 
may  be  obtained  by  eliminating  the  j^arameter  from  the  equa- 
tions found  in  the  first  step,  for  the  result  will  be  the  same  as 
that  obtained  by  eliminating  the  parameter  from  the  equations 
found  in  the  second  step. 

2.  Find  the  locus  of  the  points  of  intersection  of  two  pei-pendicular 
tangents  to  the  ellipse  6^x2  +  a^y^  —  a'^tr  =  0. 

Solution.  First  step.  The  equation  of  a  tangent  in  terms  of  its  slope  t 
is  (Art.  78)  

(4)  y  =  tx+VaH^  +  b-. 

1 
The  slope  of  the  tangent  perpendicular  to  (4)  is By  replacing  t 

1  ^ 

in  (4)  by ,  we  find  the  equation  of  the  perpendicular  tangent  to  be 


t 


(5) 


y 


X  Wi 


+  62. 


PARAMETRIC   EQUATIONS  AND  LOCI 


225 


Second  step.   As  the  parametric  equations  are  not  required,  this  step 
may  be  omitted. 

To  eliminate  t  from  (4)  and  (5)  we  write  them  in  the  forms 

tx  —  y——  ^aH"  +  5-, 
x  +  ty=  Va2  +  IjH~. 
Squaring   these    equations,    we 
obtain 

f2a:2  _  2  toy  +  ?/2  =  aH^  +  ^^ 
a;2  +  2  txy  +  ^V  =  a2  +  ^2^2, 
Adding, 

(1  +  /2)X2  +  (1+^2)^2 

=  (l  +  r^)a2  +  (l  +  f2)?,2_ 

Dividing  by  1  +  t^,  the  required 
equation  is 

x2  +  2/2  =  a2  +  62. 

The  locus  is  therefore  a  circle  whose  center  is  the  center  of  the  ellipse, 
and  whose  radius  is  VoM-"^.    It  is  called  the  director  circle. 


/  PROBLEMS 

*   1.  Find  the  locus  of  the  intersections  of  perpendicular  tangents  to  (a) 
the  parabola,  (b)  the  hyperbola  (IV),  p.  167. 

Ans.    (a)  The  directrix ;  (b)  x2  -\-  y^  =  (jfi  —  W^. 

2.  Find  the  locus  of  the  point  of  intersec- 
tion of  a  tangent  to  (a)  an  ellipse,  (b)  a  pa- 
rabola, (c)  a  hyperbola  with  the  line  drawn 
through  a  focus  perpendicular  to  the  tangent. 

Ans.  (a)  x2  +  2/2  =  a2  ;  (b)  x  =  0; 
,  (c)  x2  +  ?/2  =  a2. 

3.  Find  the  locus  of  the  point  of  intersec- 
tion of  a  tangent  to  an  equilateral  hyperbola 
and  the  line  drawn  through  the  center  per- 
pendicular to  that  tangent. 

Ans.   The  lemniscate  (x2  +  y-)- 

=  a2  (x2  -  2/2)  (Ex.  2,  Art.  46). 

4.  Find  the  locus  of  the  point  of  intersection  of  a  tangent  to  the  circle 
a;*^  +  ?/"  +  2  ox  +  a2  _  ^2  _  0  j^j^^j  ^jjg  ijj|g  drawn  through  the  origin  per- 
pendicular to  it. 

Ans.   The  limapon  (x2  +  ?/2  -|-  ax)2  =  62  ^^^2.  ^  ^2)  (problem  3,  p.  222). 


226 


NEW  ANALYTIC  GEOMETRY 


5.  Find   the   locus   of  the   foot   of   the   perpendicular   drawn   from 

the  origin  to  a  tangent  to  the  parabola 

v^  +  4  ax  +  4  a^  =0. 

a  +  X 

Ans.   The  strophoid  t/^  =  x^ (see 

«         ,  a  —  X 

figure). 

6.  Find  the  locus  of  the  intersection 
of  the  normals  drawn  at  points  on  the 

ellipse ! =  1   and  major  auxiliary 

a^      b'^ 

circle  x~  -\-  y'^  =  a^  which  have  the  same 

abscissas.     Ans.    Circle  x^  +  y~  =  {a  +  b)". 

7.  In  the  figure,  LM  is  any  half  chord 
of  the  circle  parallel  to  the  diameter  AB. 
Find  the  locus  of  P,  the  intersection  of  BL 
and  OM.  Ans.    Para})ola  xf- —  (^  —  2  ax. 

8.  A  tangent  to  the  ellipse  b-x^  +  a^y'^  =  a-b^ 
meets  the  axes  of  x  and  y  in  A  and  B  re- 
spectively. From  A  draw  a  line  1|  to  01",  and 
from  B  a  line  II  to  OX.  What  is  the  locus 
of  their  point  of  intersection  ? 
Ans.  x^y^-a^y^  +  b^x"^.  (Problem  1,(6),  p.  210.) 

9.  Work  out  Problem  8  when  the  ellipse  is 
replaced  by  a  hyperbola. 

A  somewliat  different  class  of  locus  problems  is  illustrated 
in  the  following  example. 

EXAMPLE 

What  is  the  locus  of  the  middle  points  of  a  system  of  parallel  chords 
of  an  ellipse  ? 

Solution.    Let  the  equation  of  the 
system  of  parallel  chords  be 
(0)  y  =  mx  +  k, 

w'here  A;  is  a  parameter  and  m  =  slope 
of  chords.    Let  the  value  of  k  for  the 
chord  P1P2  be  A'j ;  that  is, 
(7)  y  =  mx  +  ki 

is  the  equation  of  P^Po-    Assume  that 
the  coordinates  of  P^  are  (x^,  y^),  and  of  P^  (x,^,  y^) 


PARAMETRIC  EQUATIONS  AND  LOCI 


227 


If  P'  {x\  y')  is  the  middle  point  of  P^P^,  then 

(8)  X'  =  \  (x^  +  X2),    y'  =1  {Vi  +  2/2)- 

Since  (Xj,  2/^)  and  (Xg,  y^  are  the  points  of  intersection  of  the  chord 
(7)  and  the  ellipse,  we  shall  find  their  values  by  solving 

(9)  y  =  mx-\-k-^^     and    b'^x'^  +  a?y^  =  a^h"-. 
Eliminating  ?/,  we  obtain  the  equation 

(10)  (a2»i2  +  62)  x2  +  2  aP-k^mx,  +  a^l^  -  u-b-  =  0. 


The  roots  of  this  equation  are  x-^  and  x.,,  and,  from  (8),  x'  equals  one 
half  the  sum  of  these  roots.  Hence  we  need  to  know  in  (10)  only  the  sum 
of  the  roots.  But,  by  algebra,* 
(11)  x^  +  X2  =  - 

Hence,  from  (8), 
(1-2)   •  x^  =  -    /^^^     _k. 

Since  (x',  y')  satisfy  (7), 
(13)  y'  =  ?nx'  +  ki=      """"  7^  +  k^  =  —^ :  k. 


2a'^k^m 
a~m^  +  b' 

a?7n 
a-m'^  +  6'- 


•  a^ni^kj^ 


b- 


dhifi  +  6- 


a2m2  +  b 
Eliminating  A:^,  from  (12)  and  (13), 

(14)  b-x'  +  ahny'  =  0. 
Dropping  the  accents  gives  the  equation  of  the  locus, 

(15)  b'^x  +  a^my  =  0. 
The  locus  is  the  straight  line  DD'  in  the  figure. 

n  c 

*In  the  quadratic  Ax'^  +Dx+  C'  =  0,  .sunt.  0/ roofs  =  -  -- ;  product  of  roots  =  — 


228  NEW  ANALYTIC  GEOMETRY 

In  a  circle  a  diameter  may  be  defined  as  the  locus  of  the  middle  points 
of  a  series  of  parallel  chords.  The  corresponding  locus  for  a  conic  section 
is  also  called  a  diameter  of  the  conic. 

Hence  we  have  the 

Theorem.    The  diameter  of  the  ellipse 

7  2    2     f         1    ''  •'>I2 

b  X   -\-  a  ij   =  (lb 
K^liich  bisects  all  chords  witli  the  slope  m  is 
b^x-\-a^my  =  Q. 

In  like  manner  (see  the  figures  on  p.  227)  we  may  prove  the 

Theorem.  TJie  diameter  wliich  bisects  all  chords  with  the 
slope  VI  of  the 

hijperbola  lrx~  —  a'^ij^  =  crlr  is  IP'X  —  cp-my  =  0  ; 

jiarabola  y'^lpxis  myz=.p. 

Every  line  through  the  center  of  an  ellipse  or  hyperbola  is  a  diameter, 
while  in  a  f»arabola  every  line  parallel  to  the  axis  is  a  diameter. 

PROBLEMS 

1.  Find  the  equation  of  the  diameter  of  each  of  the  following  conies 
which  bisects  the  chords  with  the  given  slope  m. 

(a)  X-  —  4  y2  _  16^  ,^  _  2.  Alls,   x  —  8  ?/  =  0.     . 

A,    (b)  y-  =  4x,  7/1  =  —  |.  ^ns.   y  +  4  =  0. 

/,^(c)  xy  =  6,  m  =  3.  Ans.   2/  +  3x  =  0. 

(d)  x2  + X2/  — 8  =  0,  m=— 3.  Ans.   x  —  tj  -  0. 

(e)  X'^  —  42/2  +  4X  — 16  =  0,         m  =  -  1.  Ans.   x  +  4  y  +  2  =  0. 
(f)xy  +  2if-4x  —  2y  +  6  =  0,m=^.  Ans.    2x  +  1]  ?/ —  IG  =  0. 

2.  Find  the  equation  of  that  diameter  of 

(a)  4x2  +  9?/2  =  36  passing  through  (3,  2).  Ans.  2x  —  3y  =  0. 

(b)  ?/2  =  4x  passing  through  (2,  1).  Ans.  y  =  1. 

(c)  xy  =  8  passing  through  (—  2,  3).  Ans.  3x  +  2^  —  0. 

(d)  x2  -  4  ?/  +  6  =  0  passing  through  (3,  -  4).     Ans.  x  =  3. 

(e)  x?/— (/2  +  2x  — 4  =  Opassing  through  (5.  2).   Ans.  4x  — 9?/— 2  =  0. 


PARAMETRIC   EQUATIONS  AND  LOCI  229 

3.  Find  the  equation  of  the  chord  of  the  locus  of 

A'      (a)  x^  +  y'^  =  25  which  is  bisected  at  the  point  (2,  1). 

Ans.    2x  +  2/  —  5  =  0. 

(b)  4x-^  —  (/"^  =  9  which  is  bisected  at  the  point  (4,  2). 

Ans.    8x  — 2/  — 30  =  0. 

(c)  x;/ =  4  which  is  bisected  at  the  point  (5,  3).  Ayis.  3x  +  5?/  — 30=0. 

(d)  X-  —  xy  —  8  =  0  which  is  bisected  at  the  point  (4,  0). 

»  Ans.    2x  — 2/  — 8  =  0. 

4.  Show  that  if  two  lines  thi'ough  the  center  of  the  ellipse 

6-x2  +  ay-  =  a%" 

have  slopes  m  and  ?n'  such  that  mm'  = -,  then  each  line  bisects  all 

chords  parallel  to  the  other.  " 

Draw  two  such  lines.   They  are  called  conjugate  diameters. 

5.  Through  the  point  (Xq,  t/q)  on  the  ellipse  b-x^  +  a'^y-  =  aW  a  diam- 
eter is  drawn  ;  prove  that  the  coordinates  of   the   extremities  of  its 

conjugate  diameter  are  x  =  ±  — - ,    y  =  :f  — L' . 

b  a 

6.  If  a'  and  b'  are  the  lengths  of  two  conjugate  semidiameters  of  the 
ellipse,  prove  that  a''^  +  b"-  =  a'  +  6-  (use  Example  5). 

7.  Prove  that  the  tangent  at  any  point  of  the  ellipse  is  parallel  to  the 
diameter  which  is  conjugate  to  the  diameter  through  the  given  point; 
and  hence  that  the  tangents  at  the  extremities  of  two  conjugate  diameters 
form  a  parallelogram. 

8.  Pi'ove  that  the  area  of  the  parallelogram  formed  by  the  tangents  at 
the  extremities  of  two  conjugate  diameters  of  an  ellipse  is  constant  and 
is  equal  to  4  ah. 

Hint.   The  area  in  question  is  eight  times  the  area  of  the  triangle  whose 

vertices  are  (0,  0),  (a:o.  ?/o).  and  [^ '   -  ~-^)  (see  Example  5). 

9.  Two  tangents  with  the  slopes  m^  and  m.,  are  drawn  from  a  point  P 
to  an  ellipse  6'-x-  +  a-y-  =  a-h"^.    Find  the  locus  of  P 

(a)  when  m^  +  m.^  =  0.  Ans.   x  =  0  and  y  =  0. 

\y'  '(b)  when  m^  +  m._,  =  1.  Ans.   x^  —  2xy  —  a-  =  0. 

(c)  when  m^m.^  =  1.  Ans.   x^  —  y^  —  a^  —  b^. 


CHAPTER  XIII 


CARTESIAN  COORDINATES  IN  SPACE 


86.  Cartesian  coordinates.  The  foundation  of  plane  analytic 
geometry  depends  upon  the  possibility  of  determining  a  point 
in  the  plane  by  a  pair  of  real  numbers  (x,  y).  The  study  of 
solid  analytic  geometry  is 
based  on  the  determination 
of  a  point  in  space  by  a  set 
of  three  real  numbers  x,  y, 
and  z.  This  determination  is 
accomplished  as  follows  : 

Let  there  be  given  three 
mutually  perpendicular  planes 
intersecting  in  the  lines  A'A'', 
YY',  and  ZZ',  which  will  also 
be  mutually  perpendicular. 
These  three  planes  are  called 
the  coordinate  planes  and  may  be  distinguished  as  the  A'F-plane, 
the  yZ-plane,  and  the  ZA'-plane.  Their  lines  of  intersection 
are  called  the  axes  of  coordinates,  and  the  positive  directions  on 
them  are  indicated  by  the  arrowheads.*  The  point  of  inter- 
section of  the  coordinate  planes  is  called  the  origin. 

Let  P  be  any  point  in  space  and  let  three  planes  be  drawn 
through  P  parallel  to  the  coordinate  planes  and  cutting  the 
axes  at  A,  B,  and  C.    These  three  planes  together  with  the 

*  A'A^and  ZZ'  are  supposed  to  be  in  the  plane  of  the  paper,  the  positive 
direction  on  XX'  being  to  tlie  right,  tliat  on  ZZ'  being  upioard.  YY'  is  sup- 
posed to  be  perpendicular  to  the  plane  of  the  paper,  the  positive  direction  be- 
ing in  front  of  the  paper,  that  is,  from  the  plane  of  the  paper  toward  the  reader 

230 


CARTESIAN  COORDINATES  IN  SPACE  231 

coordinate  planes  form  a  rectangular  parallelepiped,  of  which 
P  and  the  origin  O  are  opposite  vertices,  as  in  the  figure. 
The  three  edges  OA  =  x,  OB  =  y,  and  OC  =  z  are  called  the 
rectangular  coordinates  of  P. 

Any  point  P  in  space  determines  three  numbers,  the  coordi- 
nates of  P.  Conversely,  given  any  three  real  numbers  x,  y,  and  z, 
a  point  P  in  space  may  always  be  constructed  whose  coordinates 
are  x,  y,  and  z.  For  if  we  lay  off  OA  =  x,  OB  =  y,  and  OC  =  z, 
and  draw  planes  through  A,  B,  and  C  parallel  to  the  coordinate 
planes,  they  will  intersect  in  a  point  P.    Hence 

Every  2^0  int  determines  three  real  numbers,  and  conversely, 
three  real  numbers  determine  a  point. 

The  coordinates  of  P  are  written  (x,  y,  z),  and  the  symbol 
P  (x,  y,  z)  is  to  be  read,  "  The  point  P  whose  coordinates  are 
X,  y,  and  ^." 

From  the  figure  we  have  the  relations 


AP  ^  OS  =  ■V(0By  -\-(ocy', 

BP=OR^  ■V{OC)--h(OAy ; 
CP  =  OQ  =  V((9J)-  +  (0£/; 

OP  =  V(a4  y  +  (pBf  +  {ocf. 

Hence,  letP  (cc,  y,  z)  be  any  point  in  space;  then  its  distance 
from  the  A'  F-plane  is  z, 
from  the  FZ-plane  is  x, 
from  the  Z^J^-plane  is  y, 


from  the  A-axis  is  Vy^  +  z^, 


from  the  F-axis  is  V,?-  +  x^, 


from  the  Z-axis  is  Vx-'^  -|-  //-, 


from  the  origin  is  Va;^  +  y"^  +  «^- 


232 


NEW  ANALYTIC   GEOMETRY 


The  coordinate  planes  divide  all  space  into  eight  parts  called 
octants,  designated  by  0-XYZ,  0-X'YZ,  etc.  The  signs  of  the 
coordinates  of  a  point  in  any  octant  may  be  determined  by  the 

Rule  for  signs. 

X  is  positive  or  negative  accord- 
ing as  P  lies  to  the  right  or  left 
of  the  YZ-plane. 

y  is  positive  or  negative  accord- 
ing as  P  lies  in  front  or  in  back 
of  the  ZX-pIane. 

z  is  positive  or  negative  accord- 
ing as  P  lies  above  or  belotv  the 
X  Y-plane. 

Points  in  space  may  be  con- 
veniently plotted  by  marking  the 
same  scale  on  A'A''  and  ZZ'  and 
a  somewhat  smaller  scale  on  YY'.  Then  to  plot  any  point,  for 
example  (7,  6,  10),  we  lay  off  OA  =  7  on  OX,  draw  AQ  jjarallel 
to  0  Y  and  equal  to  6  units  on  O  Y,  and  QP  parallel  to  OZ  and  equal 
to  10  units  on  OZ. 

PROBLEMS 

1.  What  are  the  coordinates  of  the  origin  ? 

2.  Plot  the  following  sets  of  points  : 

(a)  (8,  0,2),  (-3,  4,  7),  (0,0,  5). 

(b)  (4,  -  3,  6),  (-  4,  6,  0),  (0,  8,  0). 

(c)  (10,  3,  -  4),  (-  4,  0,  0),  (0,  8,  4). 

(cl)  (3,  -  4,  -  8),  (-  5,  -  6,  4).  (8,  6,  0). 

(e)  (-4,  -8,  -0),  (3,  0,7),  (6,  -4,2). 

(f)  (-6,4,  -  4),  (0,  -  4,  6),  (9,  7,  -2). 

3.  Calculate  the  distances  of  each  of  the  following  points  to  each  of 
the  coordinate  planes  and  axes  and  to  the  origin  : 

(a)  (2,  -  2,  1),     (b)  (3,  -  4,  -  .3),     (c)  /-i,  -  1,  ^ 

4.  Show  that  the  following  points  lie  on  a  sphere  whose  center  is  the 
origin  and  whose  radius  is  3  : 

(VS,  -  2,  V2),     (2  a/2.  0,-1),     (-2,  2,  1),     (-  Vs,  Vs,  1). 


CARTESIAN  COORDINATES  IN  SPACE  233 

5.  Show  that  the  following  points  He  on  a  circular  cylinder  of  radius 
5  whose, axis  is  the  F-axis  : 

(3,  -  8,  4),  (2  VI,  6,  V5),  (-  4,  0,  -  3),  (1,  J,  2  V6). 

6.  Where  can  a  point  move  ifx  =  0?   ii  y  =  0?   it  z  =  0? 

7.  Where  can  a  point  move  if  a;  =  0  and  y  =  0?  ii  y  =  0  and  2  =  0? 
if  z  =  0  and  x  =  0  ? 

8.  Show  that  the  points  (x,  y,  z)  and  (—  x,  ?/,  z)  are  symmetrical  with 
respect  to  the  FZ-plane ;  (x,  y,  z)  and  (x,  —  y,  z)  with  respect  to  the  ZX- 
plane  ;  (x,  y,  z)  and  (x,  y,  —  z)  with  respect  to  the  A''Y'-plane. 

9.  Show  that  the  points  (x,  y,  z)  and  (—  x,  —  y,  z)  are  symmetrical  with 
respect  to  ZZ';  (x,  y,  z)  and  (x,  —  y,  —  z)  with  respect  to  XX';  (x,  y,  z) 
and  (—  X,  ?/,  —  z)  with  respect  to  YY';  (x,  ?/,  z)  and  (—  x,  —  ?/,  —  z)  with 
respect  to  the  origin. 

10.  What  is  the  value  of  z  if  P  (x,  y,  z)  is  in  the  XY-plane  ?  of  x  if  P 
is  in  the  FZ-plane  ?   oi  y  ii  P  is  in  the  ZX-plane  ? 

11.  What  are  the  values  of  y  and  z  if  P  (x,  ?/,  z)  is  on  the  X-axis  ?  of 
z  and  X  if  P  is  on  the  F-axis  ?   of  x  and  y  ii  P  is  on  the  Z-axis  ? 

12.  A  rectangular  parallelepiped  lies  in  the  octant  0-XYZ  with  three 
faces  in  the  coordinate  planes.  If  its  dimensions  are  a,  6,  and  c,  what  are 
the  coordinates  of  its  vertices  ? 

87.  Orthogonal  projections.  To  extend  the  first  theorem  of 
projection,  Art.  31,  we  define  the  angle  between  two  directed  lines 
in  space  which  do  not  intersect  to  be  the  angle  between  two 
intersecting  directed  lines  drawn  parallel  to  the  given  lines 
and  having  their  positive  directions  agreeing  with  those  of  the 
given  lines. 

The  definitions  of  the  orthogonal  projection  of  a  point  upon 
a  line  and  of  a  directed  length  AB  upon  a  directed  line  hold 
when  the  points  and  lines  lie  in  space  instead  of  in  the  plane. 
It  is  evident  that  the  projection  of  a  point  upon  a  line  may 
also  be  regarded  as  the  point  of  intersection  of  the  line  and 
the  plane  passed  through  the  point  perpendicular  to  the  line. 
As  two  parallel  planes  are  equidistant,  then  the  projections  of 
a  directed  length  A  B  upon  two  parallel  lines  whose  positive  direc- 
tions agree  are  equal. 


234 


NEW  ANALYTIC   GEOMETRY 


A 

J/ 

/. 

i> 

/ 

^'1 

/ 



y 

c' 

A^^ 

\-'1f^ 

1 

^p' 

c 

1 

,D 

X 

/ 

Q 

/ 

First  Theorem  of  Projection.    If  A   and  B  are  points 
■upon  a  directed  line  making  an  angle  of  y  with  a  directed  line 
CD,  then  the 
(I)  projection  of  the  length  AB  upon  CD  =  AB  cos  y. 

P7vof.  Draw  CD'  through  A  parallel  to  CD.  Then,  by  defi- 
Bition,  the  angle  between  AB  and  CD'  equals  y.  Since  CD'  and 
AB  intersect  we  may  apply  the 
first  theorem  of  projection  in 
the  plane,  and  hence  the 

projection  of  the  length  AB 

upon  C'D'  =  AB  cos  y. 
Since  the  projection  of  siB  on 
CD  equals  the  projection  of  AB 
upon  CD'  we  get  (I).       q.e.d. 

Second  Theorem  of  Projection.  If  each  segment  of  a 
broken  line  in  space  he  given  the  direction  determined  vn pcbssing 
continuously  from  one  extremXtij  to  the  other,  then  the  algebraic 
sum,  of  the  projections  of  the  segments  iipjon  any  directed  line 
equals  the  projection  of  the  closing  line. 

The  proof  given  on  page  69  holds  whether  the  broken  line 
lies  in  the  plane  or  in  space. 

Corollary  I.  The  projections  of  the  line  joining  the  origin  to 
any  point  P  on  the  axes  of  coordinates  are  respect ioely  the  coordi- 
nates of  P. 

For  the  projection  of  OP  (Fig.,  p.  230)  upon  OX  equals  OA, 
since  A  is  the  })rojection  of  P  on  OX.  Similarly  for  the  pro- 
jections on  OY  and  OZ. 

Corollary  II.     Given  any  two  points  P^   (x^,  y^,  z^  and  P^ 

(^2'  y-v  -2)'  ^^^en 

jTj  —  JTi  =  projection  of  P^P^  upon  XX', 

y^—  y^=  projection  of  P^Pz  upon  YY\ 

z^—  z^=  projection  of  P^P^  ^po'^  ^^' • 


CARTESIAN  COORDINATES  IN  SPACE  235 

For  if  we  project  P^OP^  and  P^P.^  upon  A' A'',  we  have  the 
projection  of  P^O-'r  projection  of  0P,^=  projection  of  P^P^. 
But  by  Corollary  I, 

projection  of  P^O  —  —  x^,     projection  of  OP^  =  x^. 

.'.  x^  —  x^=  projection  of  P-^P.^  upon  A  A'. 

In  like  manner  the  other  formulas  are  proved. 

Corollary  III.    If  the  sides  of  a  jjolygon  he  given  the  direction 

established  by  jjassing  contimiously  around  tlie  jierimeter,  the 

•sum  of  the  projections  of  the  sides  upon  any  directed  line  is  zero. 

PROBLEMS 

1.  Find  the  projections  upon  each  of  the  axes  of  the  sides  of  the  tri- 
angles whose  vertices  are  the  following  points,  and  verify  the  results  by 
Corollary  III.      ^^^  ^_3^  ^^  _  g^^  ^.^  _  ^^  ^^^  ^g^  ^^  q^_ 

(b)  (-  4,  -  8,  -  G),  (3,  0,  7),  (6,  4,  -  2). 

(c)  (10,  3,  -4),  (-4,  0,2),  (0,8,  4). 

(d)  (-  0,  4,  -  4),  (0,  -  4,  G),  (9,  7,  -  2). 

2.  If  the  projections  of  PjPg  ^"  ^^^  2clQ'S,  are  respectively  3,  —  2,  and  7,. 
and  if  the  coordinates  of  P^  are  (—  4,  3,  2),  find  the  coordinates  of  P„. 

Ans.    (-  1,  1,  9). 

3.  A  broken  line  joins  continuously  the  points  (6,  0,  0),  (0,  4,  3), 
(—  4,  0,  0),  and  (0,  0,  8).  Find  the  sum  of  the  projections  of  the  segments 
and  the  projection  of  the  closing  line  on  (a)  the  AT-axis,  (b)  the  F-axis,^ 
(c)  the  Z-axis,  and  verify  the  results.    Construct  the  figure. 

4.  A  broken  line  joins  continuously  the  points  (6,  8,  —  3),  (0,  0,  —  3), 
(0,  0,  6),  (—  8,  0,  2),  and  (—  8,  4,  0).  Find  the  sum  of  the  projections  of 
the  segments  and  the  projection  of  the  closing  line  on  (a)  the  A'-axis, 
(b)  the  Y-axis,  (c)  the  Z-axis,  and  verify  the  results.   Construct  the  figure. 

5.  Find  the  projections  on  the  axes  of  the  line  joining  the  origin  to 
each  of  the  points  in  Problem  1. 

6.  Find  the  angle  between  each  axis  and  the  line  drawn  from  the 
origin  to  ^  g    ^ 

(a)  the  point  (8,  6,  0).  Aw^.    cos-i  -  ,  cos-i  -  ,  -  • 

o  o     2 

(b)  the  point  (2,  —  1,  —  2).     Ans.    cos-i-,cos-M ),cos-M )• 


236 


NEW  ANALYTIC   GEOMETRY 


7.  Find  two  expressions  for  the  projections  upon  tlie  axes  of  the  line 
drawn  from  the  origin  to  the  point  P  {x,  y,  z),  if  the  length  of  the  line  is 
p  and  the  angles  between  the  line  and  the  axes  are  a,  /3,  and  7. 

8.  Find  the  projections  of  the  coordinates  of  P  (x,  ?/,  z)  upon  the  line 
drawn  from  the  origin  to  P  if  the  angles  between  that  line  and  the  axes 
are  a,  /3,  and  y.  Ans.   x  cos  a,  y  cos/3,  2  cos  7. 

88.  Direction  cosines  of  a  line.  The  angles  a,  /3,  and  y  between 
a  directed  line  and  the  axes  of  coordinates  are  called  the  direc- 
tion angles  of  the  line. 

If  the  line  does  not  intersect  the  axes,  then  a,  /3,  and  y  are 
the  angles  between  the  axes  and  a  line  drawn  through  the  ori- 
gin parallel  to  the  given  line  and  agreeing  with  it  in  direction. 

The  cosines  of  the  direction  angles  of  a  line  are  called  the 
direction  cosines  of  the  line. 

Reversing  the  direction  of  a  line  changes  the  signs  of  the 
direction  cosines  of  the  line. 

For  reversing  the  direction  of  a  line  changes  a,  /8,  and  y  into 
TT  —  a,  TT  —  /?,  and  tt  —  y  respectively,  and  (30,  p.  3)  cos  (tt  —  x) 
=  —  cos  X. 

Theorem.    If  a,  (3,  and  y  are 
the  direction  angles  of  a  line, 
then 
(II)  cos'^a  +  cos'^yff  +  cos^y  =  1. 

That  is,  the  sum  of  the 
squares  of  the  direction  cosines 
of  a  line  is  unity. 

Proof.  Let  AB  he  2i  line 
whose  direction  angles  are  a, 
f3,  and  y.    Through  0  draw  OP 

parallel  to  A  B   and  let  OP  =  p.    By  definition  Z  XOP  =  a, 
Z  YOP  =  (3,  ZZOP  =  y.    Projecting  OP  on  the  axes, 

(1)  X  =  p  cos  a,     y  —  p  cos  /3,     z  =  p  cos  y. 


CARTESIAN  COORDINATES  IN  SPACE  237 

Projecting  OP  and  OCQP  on  OP, 
(2)  p  =  X  cos  a  -\-  y  cos  ^  +  z  cos  y. 

Substituting  from  (1)  in  (2)  and  dividing  by  p,  we  obtain 

(II).  Q.E.D. 

„      „            ^-  COS  a       cosfl       cosy    ^, 
Corollary.    If  =  — — ^  = '-  >  fAew 

(III)     cos  a  = — »     cos  p 


±  Va^  +  &2  _^  ^2  ±  Va2  +  62  _|_  c2 

c 

cos  y  = =r  • 

±  Va^  +  &2  +  c2 

That  is,  if  the  direction  cosines  of  a  line  are  proportional  to 
three  numbers,  they  are  respectively  equal  to  these  numbers  each 
divided  by  the  square  root  of  the  sum  of  their  squctres. 

For  if  r  denotes  the  common  value  of  tlie  given  ratios,  then 
(3)  cos  a  =  ar,     cos  /3  =  br,     cos  y  =  cr. 

Squaring,  adding,  and  applying  (II), 

l=7^(«2+Z,--^-}-c^). 
1 


±  Va'^  +  b'^  +  c^ 

Substituting  in  (3),  we  get  the  values  of  cos  a,  cos  ^8,  and 

cos  y  to  be  derived. 

The  important  conclusion  just  derived  may  be  thus  stated  : 
Any  three  numbers  a,  b,  and  c  determine  the  direction  of  a 

line  in  space.    This  direction  is  the  same  as  that  of  the  line 

joining  the  origin  and  the  point  (a,  b,  c). 

If  a  line  cuts  the  XF-plane,  it  will  be  directed  upward  or  downward 
according  as  cos  y  is  positive  or  negative. 

If  a  line  is  parallel  to  the  XF-plane,  cos  7  =  0,  and  it  will  be  directed 
in  front  or  in  back  of  the  ZX-plane  according  as  cos  /3  is  posjiiye  or  negative. 

If  a  line  is  parallel  to  the  A'-axis,  cos  /S  =  cos  7=0,  and  its  positive 
direction  will  agree  or  disagree  with  that  of  the  X-axis  according  as 
cos  a  =  1  or  —  1. 


238 


NEW  ANALYTIC  GEOMETRY 


These  considerations  enable  us  to  choose  the  sign  of  the  radical  in  the 
Corollary  so  that  the  positive  direction  on  the  line  shall  be  that  given  in 
advance. 

89.  Lengths. 

Theorem.     The    length    I    of    the    line  joining    two   points 

i\  {^v  Vv  ^i)  ^'^^  ^2  (^2'  y-v  ^2)  ^  ^"^^^  ^y 

(IV)  /  =  V(ar,  -  x^f  +  (y,  -  y^Y  +  (z,  -  z^\ 

Proof.    Let  the  direction  angles  of  the  line  P^P,^  ^®  ^'  A  ^-'^^  y- 
Projecting  PjP^  on  the  axes,  we  get,  by  the  first  theorem  of 

projection  and  Corollary  II,  p.  234, 

(1)      I  cos  a  =  a-.,  —  x^,      I  cos  ^  =  y_^  —  y^,     I  cos  y  =  z,^  —  z^ 
Squaring  and  adding, 

^^(cos^  a  +  cos-  ^  +  cos-  y)  =  (.T,  -  a^j)2  +  (_y.^  —  7/^)2  +  (,-<^  _  z^^ 

=  {^x  -  ^■^'  +  O/i  -  Vo)'  +  ('^'i  -  -2)'- 

Applying  (II),  and  taking  the  square  root,  we  have  (IV). 

Q.  E.D. 

Corollary.  The  direction  cosines  of  the  line  drawn  from  P^  to 
P^  are  proportional  to  the  projections  of  P^P^  on  the  axes. 


For,  from  (1), 
cos  a     _    cos  fi 


cos  y 


^2  -  ^1     2/2  -  yi     -■! 

since  each  ratio  equals 


Also 


.l'"" 


X 


the    denominators    are    the    pro- 
jections of  PJ^,^  on  the  axes. 

If  we  construct  a  rectangular 
parallelepiped  by  passing  planes  through  P^  and  P^  parallel  to 
the  coordinate  planes,  its  edges  will  be  parallel  to  the  axes  and 
equal  numerically  to  the  projections  of  P^^  upon  the  axes. 
P^^  will  be  a  diagonal  of  this  parallelepiped,  and  hence  V^ 
will  equal  the  sum  of  the  squares  of  its  three  dimensions. 
We  have  thus  a  second  method  of  deriving  (IV). 


CARTESIAN  COORDINATES  IN  SPACE  239 

PROBLEMS 

1.  Find  the  length  and  the  direction  cosines  of  the  line  drawn  from 

(a)  P,  (4,  3,  -  2)  to  P2  (-2,  1,  -  5).  Ans.    7,  -  f ,  -  |,  -  f. 

(b)  P,  (4,  7,  -  2)  to  P„  (3,  5,  -  4).  Ans.   3,  -  1,  -  f,  -  f. 

(c)  P^  (3,  -  8,  6)  to  Pg  (6,  -  4,  6).  Ans.    5,  f,  |,  0. 

2.  Find  the  direction  cosines  of  a  line  directed  upward  if  they  are 

proportional  to  (a)  3,  6,  and  2  ;  (b)  2,  1,  and  —  4  ;  (c)  1,  —  2,  and  3. 

/x362^,^2  1  4  1-2       3 

Ans.    (a)  -.-,-;  (b) —  , —, —  ;  (c)  — = , -— ^ , —^  • 

7    7    7  _  V2I    -  V21    +  V21  V14    Vl4    Vl4 

3.  Find  the  lengths  and  direction  cosines  of  the  sides  of  the  triangles 
whose  vertices  are  the  following  points  ;  then  find  the  projections  of 
the  sides  upon  the  axes  by  the  first  theorem  of  projection  and  verify 
by  Corollary  III,  p.  236. 

(a)  (0,0,3),  (4,0,0),  (8,0,0). 

(b)  (3,  2,  0),  (-  2,  5,  7),  (1,  -  3,  -  5). 

(c)  (-4,0,6),  (8,2, -1),  (2,  4,6). 

(d)  (3,  -  3,  -  3),  (4,  2,  7),  (-  1,  -  2,  -  5). 

4.  In  what  octant  {0-XYZ,  0-X'YZ,  etc.)  will  the  positive  part  of 
a  line  through  0  lie  if 

(a)  cosar>0,  cos/3 >0,  cos7>0?  (e)  cos  ar<0,  cosj8>0,  cos7>0  ? 

(b)  cosa>0,  cos/3>0,  cos7<0?  (f)  cos  a-<0,  cosi3<0,  cos7>0  ? 

(c)  cosa>0,  cos/3<0,  cos7<0?  (g)  cos  a<0,  cos/3<0,  cos7<0  ? 

(d)  cos  ct  >  0,  cos  /S  <  0,  cos  7  >  0  ?  (h)  cos  a  <  0,  cos  /S  >  0,  cos  7  <  0  ? 

5.  What  is  the  direction  of  a  line  if  cos  a  =  0?  cos  /3  =  0?  cos  7  =  0? 
cos  a  =  cos  p  —  0?    cos  /3  =  cos  7  =  0?    cos  7  =  cos  or  =  0  ? 

6.  Find  the  projection  of  the  line  drawn  from  the  origin  to  P^  (5,  —  7,  6) 
upon  a  line  whose  direction  cosines  are  f,  —  f,  and  |.  Ans.    9. 

Hint.  The  projection  of  OPj  on  any  line  equals  the  projection  of  a  broken 
line  whose  segments  equal  the  coordinates  of  Pj. 

7.  Find  the  projection  of  the  line  drawn  from  the  origin  to  P^  (Xj,  ?/^,  Zj) 
upon  a  line  whose  direction  angles  are  a,  /3,  and  7. 

Ans.   a;j  cos  a  +  ?/j  cos/3  4- Zj  COS7. 

8.  Show  that  the  points  (-  3,  2,  -  7),  (2,  2,  -  3),  and  (-  3,  6,  -  2)  are 
the  vertices  of  an  isosceles  triangle. 

9.  Show  that  the  points  (4,  3,  -  4),  (-  2,  9,  -  4),  and  (-  2,  3,  2)  are 
the  vertices  of  an  equilateral  triangle. 


240  NEW  ANALYTIC  GECMETRY 

10.  Show  that  the  points  (-4,0,2),  (- 1,  3V3, 2),  (2,0,2),  and 
(—  1,  Vi,  2  +  2  Ve)  are  the  vertices  of  a  regular  tetrahedron. 

11.  What  does  formula  (IV)  become  if  P^  and  P^  lie  in  the  XF-plane  ? 
in  a  plane  parallel  to  the  ATF-plane  ? 

12.  Show  that  the  direction  cosines  of  the  lines  joining  each  of  the 
points  (4,  —  8,  6)  and  (—  2,  4,  —  3)  to  the  point  (12,  —  24,  18)  are  the  same. 
How  are  the  three  points  situated  ? 

13.  Show  by  means  of  direction  cosines  that  the  three  points  (3,  —  2,  7), 
(6,  4,  —  2),  and  (5,  2,  1)  lie  on  a  straight  line. 

14.  What  are  the  direction  cosines  of  a  line  parallel  to  the  X-axis  ?  to 
the  F-axis  ?   to  the  Z-axis  ? 

15.  What  is  the  value  of  one  of  the  direction  cosines  of  a  line  parallel 
to  the  XF-plane  '?  the  FZ-plane  ?  the  ZX-plane  ?  What  relation  exists 
between  the  other  two  ? 

16.  Show  that  the  point  (—  1,  —  2,  —  1)  is  on  the  line  joining  the  points 
(4^  _  7^  3)  and  (—  6,  3,  —  5)  and  is  equally  distant  from  them. 

TT  TT 

17.  If  two  of  the  direction  angles  of  a  line  are  —  and  — ,  what  is  the 

third?  ^  ^        TT       2  7r 

A  ns.   —  or  —  • 
3         3 

18.  Find  the  direction  angles  of  a  line  which  is  equally  inclined  to  the 
three  coordinate  axes.  Aiis.    a  =  ^  =  y  =  cos-i  ^Vs. 

19.  Find  the  length  of  a  line  whose  projections  on  the  axes  are 
respectively 

(a)  6,  —  3,  and  2.  Ans.    7. 

(b)  12,  4,  and  -  3.  Ans.   13. 

(c)  —2,-1,  and  2.  Ans.    3. 

90.   Angle  between  two  directed  lines. 

Theorem,  /fa,  (3,  y  and  a',  /3',  y'  are  the  direction  angles  of  two 
directed  lines,  then  the  angle  6  between  them  is  given  bij 

(V)  cos  9  =1  cos  a  cos  a'  +  cos  p  cos  ^'  +  cos  y  cos  y'. 

Proof.  Draw  OP  and  OP'  (figure,  p.  241)  parallel  to  the  given 
lines  and  let  OP  =  p.    Then,  by  definition, 

Z  POP'  =  0. 


CARTESIAN  COORDINATES  IN  SPACE 


241 


Now,  if  the  coordinates  of  P  are  (x,  y,  z),  then,  in  the  figure, 

OA—x,     AB  =  y,     BP  =  z. 
Project  OP  and  OABP  on  OP'.    Then 

(1)  p  cos  6  =  x  cos  a'  -\-  y  cos  /3'  +  z  cos  y'. 
Projecting  OP  on  the  axes, 

(2)  X  =  p  cos  a,     y  =  p  cos  /?, 

Substituting  in  (1)  from  (2) 
and  dividing  by  p,  we  ob- 
tain   (V).  Q.E.D. 

Theorem.  If  a,  (3,  y  and  a', 
/3',  y'  are  the  direction  angles 
of  two  lines,  then  the  lines  are 

(a)  parallel  and  in  the  same 
direction*  when  and  only  when 

a  =  a\      (3  =  P',      y  =  y'; 

(b)  perpendicular^  when  and 
only  when 

cos  a  cos  a'  -\-  cos  jB  cos  /3'  -f-  cos  y  cos  y'  =  0. 

That  is,  tivo  lines  are  parallel  and  in  the  same  direction  when 
and  only  when  their  direction  angles  are  equal,  and  perpen- 
dicular when  and  only  when  the  sum  of  the  products  of  their 
direction  cosines  is  zero. 

Proof  The  condition  for  parallelism  follows  from  the  fact 
that  both  lines  will  be  parallel  to  and  agree  in  direction  with 
the  same  line  through  the  origin  when  and  only  when  their 
direction  angles  are  equal. 

The  condition  for  perpendicularity  follows  from  (V),  for  if 

TT 

^  =  — ,  then  cos  5  =  0,  and  conversely  q.e.d. 

Li 

*  They  will  be  parallel  and  have  opposite  directions  when  and  only  when 
the  direction  angles  are  supplementary. 

t  Two  lines  in  space  are  said  to  be  perpendicular  when  the  angle  between 

them  is  -,  but  the  lines  do  not  necessarily  intersect. 


242  NEW  ANALYTIC   GEOMETRY 

In  the  applications  we  usually  have  given  not  the  direction 
cosines,  but  three  numbers  to  which  they  are  proportional. 
Hence  the  importance  of  the  following 

Corollary.  If  the  direction  cosines  of  two  lines  are  proportional 
to  a,  h,  G  and  a\  h\  c',  then^the  conditions  for  pa^'allelism  and 
pteipendicularity  are  respectively 

—  =  —  =  -;5     aa'  +  bb'  +  cc'  =  0. 
a'      b        c' 

91.  Point  of  division. 

Theorem.  The  coordinates  (x,  y,  z)  of  the  point  of  division  P 
on  the  line  joining  P.^{x^,  y^,  z^  and  P^{x^,  y.^,  z,^  such  that  the 
ratio  of  the  segvients  is         p  p 

are  given  by  the  formulas 

(VI)  x=  ^/      \    y=   \      ,%    z=   '        ' 


l+A  1+A  1+A 

This  is  proved  as  in  Art.  13. 

Corollary.   The  coordinates  (x,  y,  z)  of  the  middle  point  P  of 
the  line  Joining  P^(x^,  y^,  z^  and  P^(x,^,  y.„  s;.,)  are 

^  =  |(-^l  +  J^2).        y  =  1(^1  +1/2),        ^  =  1(^1+^2)- 

PROBLEMS 

1.  Find  the  angle   between   two   lines  whose   direction   cosines  are 
respectively 

(a)  7,  f,  -  f  and  f,  -  f,  f.  Avs.   ^. 

(b)  h-hi  and  -  J-,  j%  if.  Ai,^.    cm-^^. 

(c)  t,  -  I,  \  and  f,  f,  f .  Am.   cos-i(-  ^\). 

2.  Show  that  the  lines  whose  direction  cosines  are  f,  f,  f  ;  —  f ,  f ,  —  7  ; 
and  —  7,  I ,  f  are  mutually  perpendicular. 

3.  Show  that  the  lines  joining  the  following  pairs  of  points  are  either 
parallel  or  perpendicular. 

(a)  (3,  2,  7),  (1,  4,  6)  and  (7,  -  5,  9),  (5,  -  3,  8). 

(b)  (13,  4,  9),  (1,  7,  13)  and  (7,  16,  -  6),  (3,  4,  -  9). 

(c)  (-  6,  4,  -  3),  (1,  2,  7)  and  (8,  -  5,  10),  (15,  -  7,  20). 


CARTESIAN  COORDINATES  IN  SPACE  243 

4.  Find  the  coordinates  of  the  point  dividing  the  line  joining  the  fol- 
lowing points  in  the  ratio  given. 

(a)  (3,  4,  2),  (7,  -6,4),  \=l.  Ans.  (V,  |,  |). 

(b)  (-  1,  4,  -  6),  (2,  3,  -  7),  X  =  -  3.  Ans.  {|,  §,  -  V^). 

(c)  (8,  4,  2),  (3,  9,  6),  X  =  -  i.  Ans.  {\},  i,  0). 

(d)  (7,  3,  9),  (2,  1,  2),  X  =  4.  Am.  (3,  |,  V)- 

5.  Show  that  the  points  (7,  3,  4),  (1,  0,  6),  and  (4,  5,  —  2)  are  the  ver- 
tices of  a  right  triangle. 

6.  Show   that   the   points  (-  6,  3,   2),   (3,   -  2,   4),   (5,   7,    3),    and 
(—  13,   17,   —  1)   are   the  vertices  of   a  trapezoid. 

7.  Show  that  the  points  (3,  7,  2),  (4,  3,  1),  (1,  6,  3),  and  (2,  2,  2)  are 
the  vertices  of  a  parallelogram. 

8.  Show  that  the  points  (6,  7,  3),  (3,  11,  1),  (0,  3,  4),  and  (-  3,  7,  2) 
are  the  vertices  of  a  rectangle. 

9.  Show  that  the  points  (6,  —  6,  0),  (3,  —  4,  4),  (2,   -  9,  2),  and 
^—  1,  —  7,  6)  are  the  vertices  of  a  rhombus. 

10.  Sliowthatthepoints(7,  2,4),  (4,  —  4,2),  (9,  -  1,10),  and  (G,  -7,8) 
are  the  vertices  of  a  square. 

11.  Show  that  each  of  the  following  sets  of  points  lies  on  a  .straight 
line,  and  find  the  ratio  of  the  segments  in  which  the  third  divides  the  line 
joining  the  first  to  the  second. 

(a)  (4,  13,  3),  (3,  6,  4),  and  (2,  -  1,  5).  Ans.    -  2. 

(b)  (4,  -  5,  -  12),  (-  2,  4,  6),  and  (2,  -  2,  -  6).  Ans.    \. 

(c)  {-  3,  4,  2),  (7,  -  2,  6),  and  (2,  1,  4).  Ans.    1. 

12.  Find  the  lengths  of  the  medians  of  the  triangle  whose  vertices  are 
the  points  (3,  4,  -  2),  (7,  0,  8),  and  {-  5,  4,  6).    Ans.  VllS,  V89,  2  V29. 

13.  Show  that  the  lines  joining  the  middle  points  of  the  opposite 
sides  of  the  quadrilaterals  whose  vertices  are  the  following  points 
bisect  each  other. 

(a)  (8,  4,  2),  (0,  2,  5),  (-  3,  2,  4),  and  (8,  0,  -  6). 

(b)  (0,  0,  9),  (2,  G,  8),  (-  8,  0,  4),  and  (0,  -  8,  G). 

(C)    Pj(Xi,  V/j,  Zj),  P.i{X^,  7/2,  Zo),   P3(X3,  l/g,  Z.^,   P^{.r^,  IJ^,  Z^). 

14.  Show  that  the  lines  joining  successively  the  middle  points  of  the 
sides  of  any  quadrilateral  form  a  parallelogram. 

16.  Find  the  projection  of  the  line  drawn  from  P^  (3,  2,  —  6)  to  P^ 
(—3,  6,  —4)  upon  a  line  directed  upward  whose  direction  cosines  are 
proportional  to  2.  1,  and  —  2.  Ans.    4i. 


244  NEW  ANALYTIC   GEOMETRY 

16.  Find  the  projection  of  the  line  drawn  from  P^  (6,  3,  2)  to  P^ifi,  2,  0) 
upon  the  line  drawn. from  P^il,  —  0,  0)  to  P^i—  5,  —  2,  3).       Ans.   if. 

17.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of 
the  triangle  whose  vertices  are  (3,  0,  —  2),  (7,  —  4,  3),  and  (—  1,  4,  —  7). 

Ans.    (3,  2,  -  2). 

18.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of 
the  triangle  whose  vertices  are  any  three  points  P^,  P^,  and  Pg. 

Ans.    [}  {x^  +  x^  +  Jg),  i  (j/i  +  2/„  +  2/3),  1  {Zy  +  Z2  +  23)]- 

19.  The  three  lines  joining  the  middle  points  of  the  o^jposite  edges  of  a 
tetrahedron  pass  through  the  same  point  and  are  bisected  at  that  point. 

20.  The  four  lines  drawn  from  the  vertices  of  any  tetrahedron  to  the 
point  of  intersection  of  the  medians  of  the  opposite  face  meet  in  a  point 
which  is  three  fourths  of  the  distance  from  each  vertex  to  the  opposite 
face  (the  center  of  gravity  of  the  tetrahedron). 


CHAPTER  XIV 

SURFACES,  CURVES,  AND  EQUATIONS 

92.  Loci  in  space.  In  solid  geometry  it  is  necessary  to  con- 
sider two  kinds  of  loci : 

1.  The  locus  of  a  point  in  space  which  satisfies  one  given  con- 
dition is,  in  general,  a  surface. 

Thus  the  locus  of  a  point  at  a  given  distance  from  a  fixed 
point  is  a  sphere,  and  the  locus  of  a  point  equidistant  from  two 
fixed  points  is  the  plane  which  is  perpendicular  to  the  line  join- 
ing the  given  points  at  its  middle  point. 

2.  The  locus  of  a  point  in  space  which  satisfies  two  conditions  * 
is,  in  general,  a  curve.  For  the  locus  of  a  point  which  satisfies 
either  condition  is  a  surface,  and  hence  the  points  which  satisfy 
both  conditions  lie  on  two  surfaces,  that  is,  on  their  curve  of 
intersection. 

Thus  the  locus  of  a  point  which  is  at  a  given  distance  r  from 
a  fixed  point  7*^  and  is  equally  distant  from  two  fixed  points  P^ 
and  Pg  is  the  circle  in  which  the  sphere  whose  center  is  Pj  and 
whose  radius  is  r  intersects  the  plane  which  is  perpendicular 
to  PJ^z  at  its  middle  point. 

These  two  kinds  of  loci  must  be  carefully  distinguished. 

93.  Equation  of  a  surface.  First  fundamental  problem.  If  any 
point  P  which  lies  on  a  given  surface  be  given  the  coordinates 
(ic,  y,  z),  then  the  condition  which  defines  the  surface  as  a  locus 
will  lead  to  an  equation  involving  the  variables  x,  y,  and  ,*;. 

*  The  number  of  conditions  must  be  counted  carefully.  Thus  if  a  point  is  to 
be  equidistant  from  three  fixed  points  Pj,  P^,  and  P^,  it  satisfies  two  coiuli- 
tions,  namely,  of  being  equidistant  from  P^  and  P2  and  from  Pj  ^^^  ^^a- 

245 


246  NEW  ANALYTIC  GEOMETRY 

The  equation  of  a  surface  is  an  equation  in  the  variables  x,  y, 
and  z  representing  coordinates  such  that : 

1.  The  coordinates  of  every  point  on  the  surface  will  satisfy 
the  equation. 

2.  Every  point  whose  coordinates  satisfy  the  equation  will 
lie  upon  the  surface. 

If  the  surface  is  defined  as  the  locus  of  a  point  satisfying 
one  condition,  its  equation  may  be  found  in  many  cases  by  a 
Rule  analogous  to  that  in  Art.  17. 

EXAMPLE 

Find  the  equation  of  the  locus  of  a  point  whose  distance  from 
Pj  (3,  0,  -  2)  is  4. 

Solution.  Let  P  {x,  ?/,  z)  be  any  point  on  the  locus.  The  given  con- 
dition may  be  written  p  p  _  4 


By  (IV),  P^P  =  V(x  -  3)2+2/2  +  (2  +  2)2. 

...  V(a;  -  3)2  +  2/2  +  (.  +  2)2  =  4. 
Simplifying,  we  obtain  as  the  requii-ed  equation 

x2+2/2+z2-6x+4z-3  =  0. 
That  this  is  indeed  the  equation  of  the  locus  should  be  verified  as 
on  page  3L 

PROBLEMS 

1.  Find  the  equation  of  the  locus  of  a  point  which  is 

(a)  3  units  above  the  A'F-plane. 

(b)  4  units  to  the  right  of  the  FZ-plane. 

(c)  5  units  below  the  AT'-plane. 

(d)  10  units  back  of  the  ZA'-plane. 

(e)  7  units  to  the  left  of  the  FZ-plane. 

(f)  2  units  in  front  of  the  ZA'-plane. 

2.  Find  the  equation  of  the  plane  which  is  parallel  to 

(a)  the  A'F-pIane  and  4  units  above  it. 

(b)  the  A'l"-plane  and  5  units  below  it. 

(c)  the  ZA'-plane  and  3  units  in  front  of  it. 

(d)  the  i'Z-plane  and  7  units  to  the  left  of  it. 

(e)  the  ZA'-plane  and  2  units  back  of  it. 

(f )  the  FZ-plane  and  4  units  to  the  right  of  It. 


SURFACES,  CURVES,  AND  EQUATIONS  247 

3.  What  are  the  equations  of  the  coordinate  planes? 

4.  What  is  the  form  of  the  equation  of  a  plane  which  is  parallel  to 
the  AT-plane  ?   the  FZ-plane  ?    the  ZX-plane  ? 

5.  What  are  the  equations  of  the  faces  of  the  rectangular  parallele- 
piped which  has  one  vertex  at  the  origin,  three  edges  lying  along  the 
coordinate  axes,  and  one  vertex  at  the  point  (3,  5,  7)  ? 

6.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
point      (a)  (2,  -  2,  1)  is  3.  (d)   (-  2,  |,  0)  is  Vs. 

(b)  (0,  1,  -  2)  is  |.  (e)   (a,  6,  c)  is  d. 

(c)  (-1,3,1)  is  v'i.  (f)   {a,  13,  y)  is  r. 

7.  Find  the  efjuation  of  the  sphere  whose  center  is  the  point 

(a)  (3,  0,  4)  and  whose  radius  is  5. 

Ans.    x-  +  y-  +  z'^  —  Q  X  —  8  z  =  0. 

(b)  (—  3,  2,  1)  and  whose  radius  is  4. 

Ans.   x^  +  1/^  +  z^  +  6 X  —  iy  —  2 z  —  2  =  0.  ■ 

(c)  (6,  4,  0)  and  whose  radius  is  7. 

(d)  (a,  p,  y)  and  whose  radius  is  r. 

Ans.   x'^+  y^+  z^—2a£  —  2^y  —  2yz  +  a^+  l3-  +  y"—r^  =  0. 

8.  Find  the  equation  of  a  sphere 

(a)  having  the  line  joining  (3,  0,  7)  and  (1,  —2,-1)  for  a  diameter. 

(b)  of  radius  2,  which  is  tangent  to  all  three  coordinate  planes  in  the 
first  octant. 

(c)  of  radius  3,  which  is  tangent  to  all  three  coordinate  planes  in  the 
third  octant. 

(d)  whose  center  is  the  point  (3,  1,  —  2)  and  which  is  tangent  to  the 
XF-plane. 

(e)  whose  center  is  (6,  2,  3)  and  which  passes  through  the  origin. 

(f)  passing  through  the  four  points(-2,0,0),(0,-4,0),(0, 0,4), (8, 0,0). 

9.  Find  the  equation  of  the  locus  of  a  point  which  is  equally  distant 
from  the  points 

(a)  (3,  2, -1)  and  (4,  -3,  0).  Ans.   2x- 10?/ +  22  -  11  =  0. 

(b)  (4,  -  .3,  6)  and  (2,  -  4,  2).  Ans.    4x-l-2?/  +  8z-37=0. 

(c)  (1,  3,  2)  and  (4,  -1,  1).  Ans.    Sx- Ay  -  z  -  2  =  0. 

(d)  (4,  -  6,  -  8)  and  (-2,  7,  9).         Ans.    6x- 13?/- 17^  +  9  =  0. 
10.  Find  the  equation  of  a  plane  perpendicular  at  the  niiddle  point  to 

the  line  ioining 

*  (a)   (1,  -  2,  1)  and  (2,  -  1,  0). 

(b)  (-  3,  I,  0)  and  (0,  0,  i). 

(c)  (-  2,  1,  i)  and  (i,  0,  0). 


248  NEW  ANALYTIC  GEOMETRY' 

11.  Find  the  equations  of  the  six  planes  drawn  through  the  middle 
points  of  the  edges  of  the  tetrahedron  whose  vertices  are  the  points 
(5,  4,  0),  (2,  —  5,  —  4),  (1,  7,  —  5),  and  (—  4,  3,  4),  which  are  perpendicular 
to  the  respective  edges,  and  show  that  they  all  pass  through  the  point 
(-1,1,-2). 

12.  Find  the  equation  of  the  locus  of  a  point  which  is  three  times  as 
far  from  the  point  (2,  6,  8)  as  from  (4,  —  2,  4),  and  determine  the  nature 
of  the  locus  by  comparison  with  the  answer  to  Problem  7  (d). 

13.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  the  squares  of 
whose  distances  from  (1,  3,  —  2)  and  (6,  —  4,  2)  is  50,  and  determine  the 
nature  of  the  locus  by  comparison  with  the  answer  to  Problem  7  (d). 

14.  Find  the  equation  of  the  locus  of  a  point  whose  distance 

(a)  from  the  A''-axis  is  3. 

(b)  from  the  F-axis  is  ^.  _ 

(c)  from  the  Z-axis  is  Vs. 

15.  i  ind  the  equation  of  a  circular  cylinder 

(a)  whose  axis  is  the  F-axis  and  whose  radius  is  2.  _ 

(b)  whose  axis  is  the  Z-axis  and  whose  radius  is  Vs. 

(c)  whose  axis  is  the  A'-axis  and  whose  diameter  is  V7. 

16.  A  point  moves  so  that  the  sum  of  its  distances  to  the  two  fixed 
points  (V3,  0,  0)  and  (—  V3,  0,  0)  is  always  equal  to  4.  Find  the  equa- 
tion of  its  locus.  Ans.   x^  +  i  z'^  +  i  y-  —  4  =  0. 

17.  Find  the  equation  of  the  locus  of  a  point 

(a)  whose  distance  from  the  point  (1,  0,  0)  equals  its  distance  from 
the  rZ-plane.  Ans.    y^  +  z-— 2x  +  1  =  0. 

(b)  wdiose  distance  from  the  point  (1,  0,  0)  equals  its  distance  from 
the  Z-axis.  Ans.   z^—  2x  +  1  =  0. 

(c)  whose  distance  from  the  A'^-axis  is  one  half  of  its  distance  from  the 
FZ-plane.  Arts.    4  y"^  +  i  z"  —  x^  =  0. 

(d)  whose  distance  from  the  Z-axis  is  twice  its  distance  from  the  F-axis. 

(e)  whose  distance  from  the  origin  equals  the  sum  of  its  distances 
from  the  A"Z-plane  and  the  FZ-plane.  Ans.   z"^  —  2  xy  =  0. 

(f )  the  sum  of  whose  distances  from  the  three  coordinate  planes  is 
constant. 

(g)  whose  distance  from  the  origin  equals  the  sum  of  its  distances 
from  the  three  coordinate  planes.  Ans.   xy  +  yz  +  zx  =  0. 

(h)  whose  distance  from  the  X-axis  is  half  the  difference  of  its  dis- 
tances from  the  A'F-plane  and  the  A'Z-plane. 


SURFACES,  CURVES,  AND  EQUATIONS  249 

(i)  whose  distance  from  the  point  (0,  0,  1)  equals  its  distance  from  the 
A'F-plane  increased  by  1. 

(j)  whose  distance  from  the  Z-axis  equals  its  distance  from  the 
point  (1,  1,  0). 

18.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  whose  dis- 
tances from  the  A'-axis  and  the  Y"-axis  is  unity. 

19.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  whose  dis- 
tances from  the  three  coordinate  axes  is  unity. 

94.  Planes  parallel  to  the  coordinate  planes.  We  may  easily 
prove  the 

Theorem.    The  equation  of  a  plane  which  is  ' 

parallel  to  the  XY-plane  has  the  form     z  =  constant; 

parallel  to  the  YZ-plane  has  the  form     x  =  constant; 

parallel  to  the  ZX-plane  has  the  form     y  =  constant. 

95.  Equations  of  a  curve.    First  fundamental  problem.    If  any 

point  P  wliich  lies  on  a  given  curve  be  given  the  coordinates 
(.T,  y,  «),  then  the  two  conditions  which  define  the  curve  as  a 
locus  will  lead  to  two  equations  involving  the  variables  x,  y, 
and  z. 

The  equations  of  a  curve  are  two  equations  in  the  variables 
X,  y,  and  z  representing  coordinates  such  that : 

1.  The  coordinates  of  every  point  on  the  curve  will  satisfy 
both  equations. 

2.  Every  point  whose  coordinates  satisfy  both  equations  will 
lie  on  the  curve. 

If  the  curve  is  defined  as  the  locus  of  a  point  satisfying  two 
conditions,  the  equations  of  the  surfaces  defined  by  each  condi- 
tion separately  may  be  found  in  many  cases  by  a  Rule  anal- 
ogous to  that  of  Art.  17.  These  equations  will  be  the  equations 
of  the  curve. 

It  will  appear  later  that  the  equations  of  the  same  curve 
may  have  an  endless  variety  of  forms. 


250 


NEW  ANALYTIC  GEOMETRY 


EXAMPLES 

1.  Find  the  equations  of  the  locus  of  a  point  whose  distance  from  the 
origin  is  4  and  which  is  equally  distant  from  the  points  Pj  (8,  0,  0)  and 
Po{0,  8,  0). 

Solution.    Let  V  (x,  ?/,  z)  be  any 

point  on  the  locus. 

The  given  conditions  are 

(1)     F0  =  4,    rr^  =  pp„. 
By  (IV),         

PO  =  Vx2  +  2/-i  +  z\ 

PP.   =  V(X  -  8)-   +  2/2  +  2-2, 


Substituting  in  (1),  we  get 
Vx'-  +  y~  +  z^  =  4, 


V(X  -  8)2  +  2/2  +  22  =   Vx2  +  (2/  -  8)2  +  Z^. 

Squaring  and  reducing,  we  have  the  required  equations,  namely, 

x2  +  2/2  +  -2  =  16,     X  —  2/  =  0. 
These  equations  should  be  verified  as  in  Art.  16. 

2.  Find  the  equations  of  the  circle  lying  in  the  A'T-plane  whose  center 
Is  the  origin  and  whose  radius  is  5. 

Solution.   In  plane  analytic  geometiy  the  equation  of  the  circle  is 

(2)  x2  +  2/2  =  25. 

Regarded  as  a  problem  in  solid  analytic  geometry  we  must  have  two 
equations  which  the  coordinates  of  any  point  P  (x,  y,  z)  which  lies  on  the 
circle  must  satisfy.    Since  P  lies  in  the  AT-plane, 

(3)  z  =  0. 

Hence  equations  (2)  and  (3)  together  express  that  the  point  P  lies  in  the 
A'F-plane  and  on  the  given  circle.  The  equations  of  the  circle  are  therefore 

x2  +  2/2  =  25,     z  =  0. 
The  reasoning  in  Ex.  2  is  generaL    Hence 

If  the  equation  of  a  curve  in  the  XY-pUnie  is  known,  then  the 
equations  of  that  curve  regarded  as  a  curve  in  sjjace  are  the  given 
equation  and  s  =  0. 


SURFACES,  CURVES,  AND  EQUATIONS  251 

An  analogous  statement  evidently  applies  to  the  equations  of 
a  curve  lying  in  one  of  the  other  coordinate  planes. 
From  Art.  94  we  have  at  once  the 
Theorem.     The  equations  of  a  line  which  is  parallel  to 

the  X-axis  have  the  form     y  =  constant,     z  =  constant; 

the  Y-axis  have  the  forni     z  =  constant,     x  =  constant; 

the  Z-axis  have  the  form     x  =  constant,     y  =  constant. 

PROBLEMS 

1.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  3  units  above  the  JTF-plane  and  4  units  to  the  riglit  of  tlie  FZ-plane, 

(b)  5  units  to  the  left  of  the  FZ-plane  and  2  units  in  front  of  the  ZX- plane. 

(c)  4  units  back  of  the  ZX-plane  and  7  units  to  the  left  of  the  FZ-plane. 

(d)  9  units  below  the  A^i'-plane  and  4  units  to  the  right  of  the  FZ-plane. 

2.  Find  the  equations  of  the  straight  line  which  is 

(a)  5  units  above  the  A'F-plane  and  2  units  in  front  of  the  ZA'-plane 

(b)  2  units  to  the  left  of  the  FZ-plane  and  8  units  below  the  ATy-plane. 

(c)  3  units  to  the  right  of  the  FZ-plane  and  5  units  from  the  Z-axis. 

(d)  13  units  from  the  X-axis  and  5  units  back  of  the  ZA'-plane. 

(e)  parallel  to  the  F-axis  and  passing  through  (3,  7,  —  5). 

(f)  parallel  to  the  Z-axis  and  passing  through  (—  4,  7,  6). 

3.  What  are  the  equations  of  the  axes  of  coordinates  ? 

4.  What  are  the  equations  of  the  edges  of  a  rectangular  parallelepiped 
whose  dimensions  are  a,  6,  and  c,  if  three  of  its  faces  coincide  with  the 
coordinate  planes  and  one  vertex  lies  in  0-XYZ  ?  in  0-XY'Z  ?  in 
O-A'T'Z  ? 

5.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  5  units  from  the  origin  and  3  units  above  the  XF-plane. 

(b)  5  units  from  the  origin  and  3  units  from  the  X-axis. 

(c)  6  units  from  the  F-axis  and  3  units  behind  the  XZ-plane. 

(d)  7  units  from  the  Z-axis  and  2  units  below  the  XZ-plane. 

6.  Find  the  ecpiations  of  a  circle  defined  as  follows : 

(a)  center  on  the  Z-axis,  radius  4,  and  lying  in  the  XF-plane. 

(b)  center  on  the  X-axis,  radius  7,  and  lying  in  a  plane  parallel  to  the 
yZ-plane  and  3  units  to  the  right  of  it. 

(c)  center  on  the  Y-axis,  radius  2,  and  lying  in  a  plane  2  units  behind 
the  XZ-plane. 


252  NEW  ANALYTIC  GEOMETRY 

(d)  center  at  the  point  (1,  0,  1),  parallel  to  the  JTY-plane,  and  cutting 
the  Z-axis. 

7.  The  following  equations  are  the  equations  of  curves  lying  in  one  of 
the  coordinate  planes.  What  are  the  equations  of  the  same  curves  regarded 
as  curves  in  space  ? 

(a)  y^  =  4x.  (e)  a;^  +  42  +  6x  =  0. 

(b)  x2  +  2=2  =  16.  (f )  2/2  -  z2  _  4  2/  =  0. 

(c)  8x2  _  2/2  =  64.  (g)  yz2  ^  z^-6y  =  0. 

(d)  4^2  +  92/2  =  36.  (h)  z^-ix^  +  8z  =  0. 

8.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  5  units  above  the  Xy-plane  and  3  units  from  (3,  7,  1). 

Ans.   2  =  5,  x2  +  2/2  +  z2  —  6  X  —  14 ?/  —  2 z  +  50  =  0. 

(b)  2  units  from  (3,  7,  6)  and  4  units  from  (2,  5,  4). 

Ans.   x2  +  2/2  +  z^  -  6x  -  14  2/  -  122  +  90  =  0, 
x2  +  2/2  +  z2  -  4 X-  10  2/  —  8  2  +  29  =  0. 

(c)  5  units  from  the  origin  and  equidistant  from  (3,  7,  2)  and 
(  _  3,  _  7,   _  2) .  Ans.    x2  +  2/2  +  ^2  —  25  =  0,  3 x  +  lij  +  2z  =0. 

(d)  equidistant  from  (3,  5,  —  4)  and  (—  7,  1,  6),  and  also  from 
(4,  -  6,  3)  and  (-  2,  8,  5). 

Ans.    5x  +  2//  —  0  2  +  9  =  0,  3x— 72/  —  2  +  8  =  0. 

(e)  equidistant  from  (2,  3,  7),  (3,  —  4,  6),  and  (4,  3,  -  2). 

Ans.    2  X  -  14  2/  -  2  2  +  1  =  0,  X  +  7  2/  -  8  2  +  16  =  0. 

9.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant 
from  the  points  (6,  4,  3)  and  (6,  4,  9),  and  also  from  (—  5,  S,  3)  and 
(—  5,  0,  3),  and  determine  the  nature  of  the  locus.       Ans.   z  =  6,  y  =  4. 

10.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant 
from  the  points  (3,  7,  —  4),  (—  5,  7,  —  4),  and  (—  5,  1,  —  4),  and  deter- 
mine the  nature  of  the  locus.  Ans.   x  =—1,  y  =  4. 

11.  Determine  the  nature  of  each  of  the  following  loci  after  finding 
their  equations.    The  moving  point  is  equidistant  from 

(a)  the  three  coordinate  planes. 

(b)  the  three  coordinate  axes.    . 

(c)  the  three  points  (1,  0,  0),  (0,  1,  0),  and  (0,  0,  1). 

(d)  the  A'F-plane,  the  Z-axis,  and  the  point  (0,  0,  1). 

(e)  the  A'F-plane,  the  A'-axis,  and  the  point  (0,  0,  1). 

(f )  the  points  (1,  0,  0),  (0,  1,  0),  and  the  Z-axis. 

(g)  the  X-axis,  the  F-axis,  and  the  point  (1,  0,  0). 
(h)  the  Z-axis,   the  XF-plane,  and  the  FZ-plane. 


SURFACES,  curvp:s,  and  p:quations         253 

96.  Locus  of  one  equation.  Second  fundamental  problem.  The 
locus  of  one  equation  in  three  variables  (one  or  two  may  be 
lacking)  representing  coordinates  in  space  is  the  surface  passing 
through  all  points  whose  coordinates  satisfy  that  equation  and 
through  such  points  only. 

The  coordinates  of  points  on  the  surface  niay  be  obtained 
as  follows : 

Solve  the  equation  for  one  of  the  variables,  say  z,  assume 
pairs  of  values  of  x  and  y,  and  compute  the  corresponding 
values  of  z. 

A  rough  model  of  the  surface  might  then  be  constructed  by 
taking  a  thin  board  for  the  AT-plane,  sticking  needles  into  it 
at  the  assumed  points  (.r,  y)  whose  lengths  are  the  computed 
values  of  z,  and  stretching  a  sheet  of  rubber  over  their 
extremities. 

97.  Locus  of  two  equations.  Second  fundamental  problem.  The 
locus  of  two  equations  in  three  variables  representing  coordinates 
in  space  is  the  curve  passing  through  all  points  whose  coordi- 
nates  satisfy  both  equations  and  through  such  points  only. 
That  is,  the '  locus  is  the  curve  of  intersection  of  the  surfaces 
defined  by  the  two  given  equations. 

The  coordinates  of  points  on  the  curve  may  be  obtained  as 
follows : 

Solve  the  equations  for  two  of  the  variables,  say  x  and  y,  in 
terms  of  the  third,  z,  assume  values  for  z,  and  compute  the 
corresponding  values  of  x  and  y. 

98.  Discussion  of  the  equations  of  a  curve.  Third  fundamental 
problem.  The  discussion  of  curves  in  elementary  analytic  geom- 
etry is  largely  confined  to  curves  which  lie  entirely  in  a  plane 
which  is  usually  parallel  to  one  of  the  coordinate  planes.  Such 
a  curve  is  defined  as  the  intersection  of  a  given  surface  with  a 
plane  parallel  to  one  of  the  coordinate  planes.  The  method  of 
determining  its  nature  is  illustrated  as  follows : 


254 


NEW  ANALYTIC   GEOMETRY 


EXAMPLE 

Determine  the  nature  of  the  curve  in  which  the  plane  z  =  4  intersects 
the  surface  whose  equation  is  ?/2  +  z2  _  4  x. 

Solution.    The  equations  of  the  curve  are,  by  definition, 
(1)  y"  +  z-^  =  4x,     z  =  4. 

Eliminate  z  by  substituting  from  the  second  equation  in  the  first. 
This  gives 
<2)  y"--4x  +  16  =  0,     2  =  4. 

Equations  (2)  are  also  the  equations  of  the  curve. 

For  every  set  of  values  of  {x,  y,  z)  which  satisfy  both  of  equations  (1)  will 
evidently  satisfy  both  of  equations  (2),  and  conversely. 

If  we  take  as  axes  in 
the  plane  z  =  4  the  lines 
(XX'  and  O'Y'  in  which 
the  plane  cuts  the  ZA'-plane 
and  the  I'Z-plane,  then  the 
equation  of  the  curve  when 
referred  to  these  axes  is 
the  first  of  equations  (2), 
namely, 
<3)    2/2 -4x  + 16  =  0. 

The  locus  of  (.3)  is  a  pa- 
rabola.  The  vertex,  in  the 
plane  z  =  4,   is  the  point  ^ 
(4,  0);  alsop  =  2. 

In  plotting  the  locus  of  (3)  in  the  plane  A'O'y' the  values  of  x  and  y 
must  be  laid  off  parallel  to  O'A''  and  O'Y'  respectively,  as  in  plotting 
oblique  coordinates  (Art.  9). 

From  the  preceding  example  we  may  state  the 

Rule  to  determine  the  nature  of  the  curve  in  which  a  plane 
parallel  to  one  of  the  coordinate  planes  cuts  a  given  surface. 

■Eliminate  the  variable  occurring  in  the  equation  of  the  plane 
frovi  the  equations  of  the  plane  and  surface.  The  result  is  the 
equation  of  the  curve  referred  to  the  lines  in  ivhich  the  given 
plane  cuts  the  other  two  coordinate  planes  as  axes.  Discuss  this 
curve  by  the  methods  of  plane  analytic  geometry. 


SURFACES,  CURVES,  AND  EQUATIONS  255 

PROBLEMS 

1.  Determine  the  nature  of  the  following  curves  and  construct  their 
loci  : 

(a)  x2  _  4y2  ^  8z,  z  =  8.  (e)  x^  +  4y^  +  9z2  =  36,  y  =  l. 

(b)  x^  +  9y-  =  9z2,  z  =  2.  (f)  x^  -  4y^  +  z^  =  25,  x  =-  3. 

(c)  x2  —  4  2/2  =  4  z,  ?/  =  -  2.  (g)  x2  —  2/2  —  4  z2  +  6  X  =  0,  X  =  2. 

(d)  x2  +  2/2  +  z2  =  25,  x  =  3.  (h)  2/2  +  z2-4x  +  8  =  0,  2/  =  4. 

2.  Construct  the  curves  in  which  each  of  the  following  surfaces  inter- 
sects the  coordinate  planes : 

(a)  x2  +  42/2  +  16z2  =  64.  (d)  x2  +  9^2  ^  joz. 

(b)  x2  +  42/2-16z2  =  64.  (e)  x^-9y'^  =  10z. 

(c)  x2  -  42/2  -  16z2  =  64.  (f)  x2  +  42/2  -  16z2  =  0. 

3.  Show  that  the  curves  of  intersection  of  each  of  the  surfaces  in 
Problem  2  with  a  system  of  planes  parallel  to  one  of  the  coordinate  planes 
are  conies  of  the  same  species  (see  Art.  70). 

4.  Determine  the  nature  of  the  intersection  of  the  surface  x2  +  y'^  + 
4  z2  =  64  with  the  plane  z  —  k.  How  does  the  curve  change  as  k  increases 
from  0  to  4  ?  from  —  4  to  0  ?  What  idea  of  the  appearance  of  the  surface 
is  thus  obtained  ? 

5.  Determine  the  nature  of  the  intersection  of  the  surface  4x  —  2  2/  =  4 
with  the  plane  y  =  k;  with  the  plane  z  =  k'.  How  does  the  intersection 
change  as  fc  or  Ar'  changes  ?  What  idea  of  the  form  of  the  surface  is 
obtained  ? 

6.  In  each  of  the  following  lind  the  equations  of  the  locus,  determine 
its  nature,  and  construct  it : 

(a)  A  point  is  5  units  from  the  origin  and  3  units  from  the  Z-axis. 

(b)  A  point  is  3  units  from  both  the  X-axis  and  the  Z-axis. 

(c)  The  distance  of  a  point  from  the  Z-axis  is  equal  to  twice  its  distance 
from  the  XF-plane  and  its  distance  from  the  origin  is  2. 

(d)  A  point  is  5  units  from  the  A'-axis  and  4  units  from  the  AZ-plane. 

(e)  A  point  is  equidistant  from  the  TZ-plane  and  the  XZ-plane  and 
its  distance  from  the  X-axis  is  7.  Ans.    An  ellipse. 

(f )  A  point  is  equidistant  from  the  Z-axis,  the  FZ-plane,  and  the  point 
(2,  0,  0).  Ans.   A  parabola. 


256  NEW  ANALYTIC  GEOMETRY 

7.  The  ratio  of  the  distances  of  a  point  to  the  Z-axis  and  the  I'-axis 
respectively  is  |.    Determine  the  nature  of  its  locus  if  it  is  also 

(a)  one  unit  above  the  XF-plane. 

(b)  one  unit  in  front  of  the  XZ-plane. 

(c)  one  unit  to  the  left  of  the  FZ-plane. 

(d)  in  the  A'Z-plane. 

(e)  equidistant  from  the  A'Z-plane  and  the  Y"Z-plane. 

(f )  in  the  plane  4x  —  3z  —  12  =  0. 

8.  Find  the  equations  of  the  locus  of  a  point  whose  distance  from  the 
point  (2,  0,  0)  is  always  equal  to  three  times  its  distance  from  the  Z-axis, 
and  whose  distance  from  the  FZ-plane  is  always  unity.  Name  and  draw 
the  locus. 

9.  Find  the  equations  of  the  locus  of  a  point  which  is  equidistant  from 
the  point  (1,  —  2,  0)  and  the  Z-axis,  and  which  is  3^  units  behind  the 
ATZ-plane.    Name  and  draw  the  locus. 

10.  Find  the  equations  of  the  locus  of  a  point  which  is  equidistant  from 
the  y-axis  and  the  A'Z-plane  and  equidistant  from  the  origin  and  the 
point  (0,  0,  —  4).    Name  and  draw  the  locus. 

99.  Discussion  of  the  equation  of  a  surface.  Third  fundamental 
problem. 

Theorem.    The  locus  of  an  algebraic  equation  jj asses  through 
the  origin  if  there  is  no  constant  term  in  the  equation. 
The  proof  is  analogous  to  that  on  page  47. 

Theorem.  If  the  locus  of  an  equation  is  unaffected  by  chang- 
ing the  sign  of  one  variable  throughout  its  equation,  then  the  locus 
is  symmetidcal  with  respect  to  the  coordinate  plane  from  which 
that  variable  is  measured. 

If  the  locus  is  unaffected  by  changing  the  signs  of  two  variables 
throughout  its  equation,  it  is  symmetrical  with  respect  to  the  axis 
along  which  the  third  variable  is  measured. 

If  the  locus  is  unaffected  by  changing  the  signs  of  all  three 
variables  throughout  its  equation,  it  is  symmetrical  with  respect 
to  the  origin. 

The  proof  is  analogous  to  that  on  page  42. 


SURFACES,  CURVES,  AND  EQUATIONS 


257 


Rule  to  find  the-  intercepts  of  a  surface  on  the  axes  of  coordinates. 
Set  each  pair  of  variables  equal  to  zero  and  solve  for  real 
values  of  the  third. 

The  curves  in  which  a  surface  intersects  the  coordinate  planes 
are  called  its  traces  on  the  coordinate  planes.  From  the  Rule, 
p.  254,  it  is  seen  that 

The  equations  of  the  traces  of  a  surface  are  obtained  by  succes- 
sively  setting  a:;  =  0,  ?/  =  0,  and  z=:0  in  the  equation  of  the  surface. 

By  these  means  we  can  determine  some  properties  of  the  sur- 
face. The  general  appearance  of  a  surface  is  determined  by  con- 
sidering the  curves  in  which  it  is  cut  by  a  system  of  planes 
parallel  to  each  of  the  coordinate  planes.  This  also  enables  us 
to  determine  whether  the  surface  is  closed  or  recedes  to  infinity. 


EXAMPLE 

Discuss  the  locus  of  the  equation  ^/^  +  z^  =  4x. 

Solution.    1.  The  surface  passes  through  the  origin  since  there  is  no 
constant  term  in  its  equation. 

2.  The  surface  is  symmetrical  with  respect  to  the  A'y-plane,  the  ZX- 
plane,  and  the  JT-axis. 

For  the  locus  of  the  given 
equation  is  unaffected  by 
changing  the  sign  of  z,  of 
y,  or  of  both  together. 

3.  It  cuts  the  axes  at  the 
origin  only. 

4.  Its  traces  are  respec- 
tively the  point-circle  y'^  4- 
z"  =  a  and  the  parabolas 
z^  =  4iX  and  y"^  =  4iX. 

5.  It  intersects  the  plane 
X  =  k  in  the  curve 

2/2  +  z2  ^  4  k. 

This  curve  is  a  circle  whose  center  is  the  origin,  that  is,  is  on  the  A'-axis, 
and  whose  radius  is2  v^  if  fc>0,  but  there  is  no  locus  if  A;<0.  Hence  the 
surface  lies  entirely  to  the  right  of  the  FZ-plane. 


258  NEW  ANALYTIC  GEOMETRY 

If  A;  increases  from  zero  to  infinity,  the  radius  of  the  circle  increases 
from  zero  to  infinity  while  the  plane  x  =  k  recedes  from  the  FZ-plane. 

The  intersection  with  a  plane  z  —  k  or  y  =  kf,  parallel  to  the  XY-  or 
the  ZA'-plane,  is  seen  to  be  a  parabola  whose  equation  is 
y"^  =  4x  —  k^     or    z^  =  ix  —  k"^. 

These  parabolas  have  the  same  value  of  p,  namely  p  =  2,  and  their  ver- 
tices recede  from  the  YZ-  or  the  ZA'-plane  as  k  ork'  increases  numerically. 

PROBLEMS 

1.  Discuss  and  draw  the  loci  of  the  following  equations: 
(a)x2  +  z2  =  4x.  (k)  x^  +  y"- -  z""  =  0. 

(b)  x2  +  ?/2  4-  4  z2  =  16.  ( 1 )  a;2  -  y^  -z^  =  9. 

(c)  x2  +  2/2  -  4 z2  =  16.  (m)  x'^  +  y^  -  z'^  +  2xy  =  0. 

(d)  6x  +  iy  +  Sz  =  12.  (w)  x  +  y  -  Gz  =  6. 

(e)  3x  +  2y  +  z  =  12.  (o)  y'^  +  z^  =  25. 

(f)  X  +  23  -  4  =  0.  (p)  x2  +  2/2  -  22  _  1  ^  0, 

(g)  x2  +  2/2  -  2 z  =  0.  (q)  x2  +  7/2  -  -2  +  1  ^  0. 
(h)  x2  +  2/2  -  2x  =  0.  (  r)  4x2  -  2/2  -  22  =  0. 
(i)  x2  +  2/2  —  4  =  0.  (r)  z^  -  X  -  y  =  0. 

{{)  y^  +  z'^-x-4  =  0.  ( t )  x2  +  2/2  -  2  2X  =  0. 

2.  Show  that  the  locus  of  Ax  +  By  +  Cz  +  D  =  0  is  a  plane  by  con- 
sidering its  traces  on  the  coordinate  planes  and  the  sections  made  by 
planes  parallel  to  one  of  the  coordinate  planes. 

3.  In  each  of  the  following  find  the  equation  of  the  locus  of  the  point 
and  draw  and  discuss  it : 

(a)  The  sum  of  the  distances  of  a  point  from  the  A'Z-plane  and  the 
FZ-plane  equals  twice  its  distance  from  the  A'F-plane  increased  by  4. 

(b)  The  square  of  its  distance  from  the  Z-axis  is  equal  to  four  times 
its  distance  from  the  A'F-plane. 

(c)  Its  distance  from  the  Z-axis  is  double  its  distance  from  the  XY- 
plane. 

(d)  Its  distance  from  the  F-axis  is  twice  the  square  root  of  its  distance 
from  the  YZ-plane. 

(e)  It  is  equally  distant  from  the  point  (2,  0,  0)  and  the  FZ-plane. 

Ans.   y^  +  z^  —  4x  +  4  -0. 

(f)  It  is  equally  distant  from  the  point  (0,  2,  0)  and  the  AT-axis. 

(g)  Its  distance  from  the  Z-axis  is  equal  to  its  distance  from  the 
FZ-plane  increased  by  2. 


SURFACES,  CURVES,  AND  EQUATIONS  259 

(h)  Its  distance  from  the  point  (0,  0,  —  2)  is  equal  to  double  its  distance 
from  the  JTZ-plane  increased  by  unity. 

(i)  Its  distance  from  the  point  (i,  0,  0)  is  equal  to  half  its  distance 
from  the  yZ-plane  diminished  by  one.     Ans.    3x-  +  4 y-  +  4 z^  —  3  =  0. 

(j)  The  product  of  the  sum  and  the  difference  of  its  distances  from 
the  A'Z-plane  and  the  FZ-plane  respectively  is  equal  to  twice  its  distance 
from  the  A'l'-plane. 

4.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
point  (0,  0,  3)  is  twice  its  distance  from  the  A'F-plane,  and  discuss  the 
locus.  Ans.   a;2  +  2/"^  —  3  z2  _  6  z  4-  9  =  0. 

5.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
point  (0,  4,  0)  is  three  fifths  its  distance  from  the  ZX-plane,  and  discuss 
the  locus.  A  ns.   25  x^  +  16y'  +  25  z^  —  200y  +  400  -  0. 


CHAPTER  XV 


THE  PLANE  AND  THE  GENERAL  EQUATION  OF  THE  FIRST 
DEGREE  IN  THREE  VARIABLES 

100.  The  normal  form  of  the  equation  of  the  plane.    Let  ABC 

be  any  plane,  and  let  ON  be  drawn  from  the  origin  perpen- 
dicular to  ABC  at  D.  Let  the  jjositive  direction  on  ON  he  from 
0  toward  N,  that  is, 
from  the  origin  to- 
ward the  plane,  and 
denote  the  directed 
length  OD  by  j)  and 
the  direction  angles 
of  ON  by  a,  j3,  and  y. 
Then  the  jjosltlon  of 
any  plane  is  deter- 
mined hy  given  posi- 
tive values  of  p,  a, 
P,  and  y. 

If  p  =  0,  the  positive  direction  on  OiV,  as  just  defined,  becomes  mean- 
ingless.   Ifp  —  0,  we  shall  suppose  that  ON  is  directed  upward,  and  hence 

TT 

cos  7  >0  since  y  <  -  ■    If  the  plane  passes  through  OZ,  then  ON  lies  in  the 
Xy-plane  and  cos  7  =  0;  in  this  case  vje  shall  suppose  ON  so  directed  that 

TT 

P<-  and  hence  cos  ^>0.    Finally,  if   the  plane  coincides  with   the 
yZ-plane,  the  positive  direction  on  ON  shall  be  that  on  OX. 

Let  us  now  solve  the  problem : 

Given  the  perpendicular  distance p  from  the  origin  to  a  plane 
and  the  direction  angles  a,  y8,  y  of  this  perpendicular,  to  find 
the  equation  of  the  plane. 

260 


THE  PLANE  261 

Solution.  Let  P(x,  y,  z)  be  any  point  on  the  given  plane 
ABC.  Draw  the  coordinates  OE  =  x,  EF  =  y,  FP  =  z  of  P. 
Project  OEFP  and  OP  on  the  line  ON.  By  the  second  theorem 
of  projection, 

projection  of  OE  +  projection  of  EF  +  projection  of  FP 
=  projection  of  OP. 

Then  by  the  first  theorem  of  projection  and  by  the  defini- 
tion of  p,  a:  cos  a  4-  y  cos  /?  +  «  cos  y  =  j9. 
Transposing,  we  obtain  the 
Theorem.    Normal  form.    The  eqiiatlon  of  a  plane  is 

(I)  jrcosa +  y  cosyff +  ^cosy  — />  =  0, 

where  p  is  the  perpendicular  distance  from  the  origin  to  theplane, 
and  a,  /3,  and  y  are  the  direction  cosines  of  tltat  2i^^p>endlcular. 

Corollary.  The  equation  of  any  plane  is  of  the  first  degree  in 
X,  y,  and  z. 

101.  The  general  equation  of  the  first  degree,  Ax -\- By -\- Cz 

-\-D  =  Q.  The  question  now  arises:  Given  an  equation  of  the 
first  degree  in  the  coordinates  x,y,z\  what  is  the  locus  ?  This 
question  is  answered  by  the 

Theorem.  The  locus  of  any  equation  of  the  first  degree  In 
X,  y,  and  z, 

(II)  Ax-\-By-\-Cz-\-D  =  Q, 
is  a  plane. 

Proof.  We  shall  prove  the  theorem  by  showing  that  (TI) 
may  be  reduced  to  the  normal  form  (I)  by  multiplying  by  a 
proper  constant.  To  determine  this  constant,  multiply  (II) 
by  /c,  which  gives 

(1)  kAx  +  IcBy  +  kCz  +  kD  =  0. 


262  NEW  ANALYTIC  GEOMETRY 

Equating  corresponding  coefficients  of  (1)  and  (I), 
(2)       kA  =  cos  a,     kB  =  cos /3.      kC  =  eosy,     kD  =  —  p. 
Squaring  the  first  three  of  equations  (2)  and  adding, 
k'^{A^-\-  Br-{-  C^)  =  cos-rt  +  cos'^yS  +  cos'y  =  1. 
1 


(3)  .-.  k 


±^A^+  E"+  C'^ 


From  the  last  of  equations  (2)  we  see  that  the  sign  of  the 
radical  must  be  opposite  to  that  of  D  in  order  that  p  shall  be 
positive. 

Substituting  from  (3)  in  (2),  we  get 

cos/5 


(4) 


±  V.-l''^+5-+  C^  ±  V.  1-  +  i?2  4-  c^ 

C  —D 


eOSy  =  .  =r;  p  =  - ,  =• 

±^A--\-B-+  C  ±^A-^+B''+C^ 


We  have  thus  determined  values  of  a,  fS,  y,  and  p  such  that 
(I)  and  (II)  have  the  same  locus.  Hence  the  locus  of  (II)  is  a 
plane.  q.e.d 

If  D  =  0,  then  p  =  0 ;  and  from  the  third  of  equations  (2)  the  sign  of 
the  radical  must  be  the  same  as  that  of  C,  since  when  p  =  0,  cos7>0.  If 
D  =  0  and  C  =  0,  then  p  =  0  and  cos  7  =  0;  and  from  the  second  of 
equations  (2)  the  sign  of  the  radical  must  be  the  same  as  that  of  B,  since 
when  p  =  Q  and  cos  7  =  0,  cos  j8  >  0. 

Equation  (II)  is  called  the  general  equation  of  the  first  degree 

in  X,  y,  and  z.    The  discussion  gives  the 

Rule  to  reduce  the  equation  of  a  plane  to  the  normal  form. 
Divide  the  equation  by  ±  V /1^+  B--\-  C',  chooslmj  the  sign  of 
the  radical  opposite  to  that  of  D. 

When  Z)  =  0,  the  sign  of  the  radical  must  be  the  same  as  that 
of  C,  the  same  as  that  of  5  if  C  :=  i*  =  0,  or  the  same  as  that  of 
A\iB=C  =  D  =  Q. 


THE  PLANE  263 

From  (4)  we  have  the  important 

Theorem.  The  coefficients  of  x,  y,  and  z  in  the  equation  of  a 
plane  are  proportional  to  the  direction  cosines  of  any  line  per- 
pendicular to  the  plane. 

From  this  theorem  and  Art.  90  we  easily  prove  the  following : 
Corollary  I.    Two  planes  whose  equations  are 

Ax  +  By-\-Cz-\-D=^Q,     A'x  +  B'y  +  C'z  -\-  D'  =  0 
are  parallel  when  and  only  when  the  coefficients  ofx,  y,  and  z  are 
proportional,  that  is,         ABC 
T'^B'^'C'' 

Corollary  II.    Two  jylanesaj'eperpendlcnlarwhen  and  only  when 
AA'+BB'+  CC'=  0. 

Corollary  III.  A  plane  whose  equation  has  the  form 

Ax  -\-  By  -\-  D  —  0  is  perpendicular  to  the  XY-pjlane  ; 

By  -\-  Cz  -\-  D  :=  ()  Is  pjerpendlcular  to  the  YZplane  ; 

Ax  -\-  Cz  -{-  D  =  0  Is  jyerjjendlcitlar  to  the  ZX-jylane. 

That  is,  if  one  variable  is  lacking,  the  pilane  is  pjerjjendlcular  to 

the  coordinate  plane  correspjondlng  to  the  two  variables  which 

occur  in  the  equation. 

For  these  planes  are  respectively  perpendicular  to  the  planes 
s:  =  0,  aj  =  0,  and  ?/  =  0  by  Corollary  II. 

Corollary  IV.    A  plane  whose  equation  has  the  form 

Ax  4-  7)  =  0  is  perpendicular  to  the  axis  ofx; 

By  -f  Z)  =  0  is  p)erpendlcular  to  the  axis  of  y  ; 

Cz  -\-  D  =  0  is  pe7pendlcular  to  the  axis  of  z. 

That  is,  if  two  variables  are  lacking,  the  pdane  Is  perpendicular 

to  the  axis  coiTespondlng  to  the  variable  which  occurs  in  the 

equation. 

For  two  of  the  direction  cosines  of  a  perpendicular  to  the 
plane  are  now  zero,  and  hence  this  line  is  parallel  to  one  of  the 
axes  and  the  plane  is  therefore  peri)endicular  to  that  axis. 


264  NEW  ANALYTIC  GEOMETRY 


PROBLEMS 

1.  Find  the  intercepts  on  the  axes  and  the  traces  on  the  coordinate 
planes  of  each  of  the  following  planes  and  construct  the  figures : 

(a)  2x  +  3y  +  4z- 24  =  0.  (e)  5x-72/-35  =  0. 

(h)  7x-Sy +  z-21  =  0.  (f)  4x  +  3z  +  36  =  0. 

(c)  9x-  7?/-9z  +  63  =  0.  (g)  5?/ -  8z  -  40  =  0. 

(d)  6x  + 4?/-z  +  12  =  0.  (h)  3x  +  5z  +  45  =  0. 

2.  What  are  the  intercepts  and  the  equations  of  the  traces  on  the  coor- 
dinate planes  of  the  plane  Ax  +  By+Cz  +  B  =  0? 

3.  Find  the  equations  of  the  planes  and  construct  them  by  drawing 
their  traces,  for  which 

(a)  ct  =  -,  /3  =  -,  7  =  -,  p  =  6.  Ans.    V2x  +  y  +  z  —  12  =  0. 

,,  s  2  TT    „      37r  ""  _  .  /—  -rt       ^ 

(b)  a  =  ^,l3  =  — ,7  =  ^,p  =  8.         Ans.   x  +  ■\2y  —  z  +  16  -  0. 

o  4  3 

,    ^    cos  a         COSjS         cos  7  .  <  /.  «         ,     o  no         r. 

(c)  = — ,  p  =  4.  Ans.   6x  —  2w  +  3z  —  28  =  0. 

^  '      6  -  2  3 

,.,   cosor      cos/3      cos 7 

(d)  —^  =  -—J  =  —^,  p  =  2.  Ans.    2x  +  y  +  2z  +  6  =  0. 

4.  Find  the  equation  of  the  plane  such  that  the  foot  of  the  perpen- 
dicular from  the  origin  to  the  plane  is  the  point 

(a)(-3,  2,  6).  Ans.    3x  -  2y  -  6z  +  49  =  0. 

(b)  (4,  3,  -  12).  Ans.    4x -F  3y  -  12z  -  169  =  0. 

(c)  (2,2,-1).  Ans.    2x  +  2y-z-9-0. 

6.  Reduce  the  following  equations  to  the  normal  form  and  find  a,  /3, 
7,  and  p  : 

(a)  6x  —  3y  +  2z-7  =  0.  Ans.   cos-i  f,  cos-i  (— f),  cos-i  f,  1. 

.         2  7r    TT    2  7r     . 

(b)  x-  V27/  +  z  +  8  =  0.     Ans.    —,-,—,4. 

(e)  2x  —  2y  —  z +  12  =  0.    Ans.    cos-i  (— |),  cos-i  |,  cos-i  ^,  4. 

{d)y-z+10  =  0.  ^ns.   -,—,-,  5  V2. 

2      4      4 

{e)  Sx  +  2y-6z  =  0.         Ans.    cos-i  (- f),  cos-i(- f),  cos-i  ^,  0. 
6.  Find  the  distance  from  the  origin  to  the  plane  12x  — 42/4-3z— 39  =  0. 


THE  PLANE  265 

7.  Find  the  area  of  the  triangle  which  the  three  coordinate  planes  cut 
from  each  of  the  following  planes : 

(a)  2  X  +  2  ?/  +  z  -  12  =  0.  Am.   54. 

(b)  6a: -2?/ -32  +  21  =  0. 

(c)  12x- 3?/  + 42-13  =  0.  r- 

(d)  X  +  5?/  +  72  -  3  =  0.  Anfi.   — ^  . 

(e)  X  -  2  2/  +  3  2  -  6  =  0.  ^'* 

(f)  dx  +  2y-z  +  18  =  0. 

Hint.  Find  the  volume  of  the  tetrahedron  formed  by  the  four  planes  by  find- 
ing the  intercepts.  Set  this  equal  to  the  product  of  the  required  area  by  one 
third  the  distance  of  the  given  plane  from  the  origin,  and  solve. 

8.  Find  the  distance  between  the  parallel  planes 6x  +  2?/  —  32  —  G3  =  0 
and  6x  +  2?/  — 32  +  49  =  0.  Ans.    16. 

9.  Find  the  equation  of  a  plane  parallel  to  the  plane  2x  +  2y  +  z  — 
15  =  0  and  two  units  nearer  to  the  origin. 

10.  Show  that  the  following  pairs  of  planes  are  either  parallel  or  per- 
pendicular : 

r2x  +  57/-62  +  8  =  0,  J6x-3y  +  2z-7  =  0, 

^''  \6x  +  15?/-182-5  =  0.  ^^'  \3x  +  2?/-62  + 28  =  0. 

r3x-  52/-  42+  7  =  0,  Ji4x-  7?/ -212-50  =  0, 


{6x  +  2y  +  2z-7  =  0.  ^  '   \2x  -  y  -  Sz  +  12  =  0. 

11.  What  may  be  said  of  the  position  of  the  plane  (I),  Art.  100,  if 
(a)  cosa=:0?  (c)  cos7  =  0?  (e)  cos/3  =  cosy  =  0  ? 

.    (b)  cos  /3  =  0  ?  (d)  cos  a  =  cos  /3  =  0  ?  (f )  cos y  =  cos  a  =  0  ? 

12.  For  what  values  of  ex,  ft  7,  and  p  will  the  locus  of  (I),  Art.  100,  be 
parallel  to  the  A'F-plane  ?  the  YZ-plane  ?  the  ZA'-plane  ?  coincide  with 
one  of  these  planes  '? 

13.  For  what  values  of  a,  /3,  7,  and  p  will  the  locus  of  (I),  Art.  100, 
pass  through  the  AT-axis  ?   the  Y-axis  ?   the  Z-axis  ? 

14.  Find  the  coordinates  of  the  point  of  intersection  of  the  planes 
x  +  2?/  +  z  =  0,  x-2?/-8  =  0,  x  +  2/  +  2-3  =  0.       Ans.    (2,  -  3,  4). 

15.  Show  that  the. plane  x  +  2y—  2z  —  9  =  0  passes  through  the  point 
of  intersection  o\  the  planes  x  +  y  +  z  — 1  =  0,  x  —  y  —  z  —  l  =  0,  and 
2x  +  3?/-8  =  0. 

16.  Show  that  the  four  planes  x  +  2/  +  2z  —  2  =  0,  x  +  y— 2z  +  2  =  0, 
X  —  ?/  +  8  =  0,  and  Sx  —  y  —  2z  +  18  =  0  pass  through  the  same  point. 

17.  Show  that  the  planes  2x  —  y  +  z  +  S  =  0,  x  —  y  +  iz  =  0,  3x  + 
2/-22  +  8  =  0,  4x-2?/  +  22-5  =  0,  9x  +  3?/- 6z- 7  =  0,  and  7x- 
7(/  +  282  —  6  =  0  bound  a  parallelepiped. 


266 


NEW  ANALYTIC   GEOMETRY 


18.  Show  that  the  planes  6x  —  3y  +  2z  =  4,3x  +  2y  — 6z=10,2x  + 
6y  +  3z  =  9,  Sx  +  2y-Gz  =  0,  \2x  +  36  ?/ +  I82  -  11  ^  0,  and  12x- 
6y  +  4z— 17  =  0  bound  a  rectangular  parallelepiped. 

19.  Show  that  the  planes a;  +  2?/  —  z  =  0,  ?/4-7z  —  2  =  0,  x  —  2y  — 
z  —  4  =  0,  2x  +  ?/  —  8  =  0,  and  3x  +  3?/  —  s  —  8  =  0  bound  a  quadrangu- 
lar pyramid. 

20.  Derive  the  conditions  for  parallelism  of  two  planes  from  the  fact 
that  two  planes  are  parallel  if  all  their  traces  are  parallel  lines. 

102.  Planes  determined  by  three  conditions.    The  equation 

(1)  Ax+B!j^rz  +  D  =  0 

represents,  as  we  know,  all  planes.  The  statement  of  a  problem, 
to  find  the  equation  of  a  certain  plane,  may  be  such  that  we  are 
able  to  write  down  three  homogeneous  equations  in  the  coeffi- 
cients A,  B,  (',  J),  which  we  can  then  solve  for  three  coefficients 
in  terms  of  the  fourth.  When  these  values  are  substituted  in 
(1),  the  fourth  coefficient  will  divide  out,  giving  the  required 
equation. 

EXAMPLES 

1.  Find  the  equation  of  the  plane  which  passes  through  the  point 
p^  (2,  —  7,  f )  and  is  parallel  to  the  plane  21  x  —  12  ?/  +  28  2  —  84  =  0. 

Solution.  Let  the  equa- 
tion of  the  required  plane  be 

(2)  Ax+  By  +  Cz  +  D  =  0. 

Since  P-y  lies  on  (2),  we 
may  substitute  a;  =  2,  ^  =  —  7, 
z-l,  giving 
(.3)  2A-1  B+  I  C  +  X>  =  0. 

Since  (2)  is  parallel  to  the 
given  plane  (Corollary  I, 
p.  263), 

^  '       21       -12      28 

Equations  (3)  and  (4)  are 
three  homogeneous  equations  in  ^,  B,  C,  D. 

Solving  (3)  and  (4)  for  A,  B,  and  D  in  terms  of  C, 
A  =  IC,     B=-  I  C,     D  =  -  6C. 


THE  PLANE 


267 


Substituting  in  (2),  ^  Cx  —  I  Cy  +  Cz  —  6  C  =  0. 

Clearing  of  fractions  and  dividing  by  C, 

21  X  — 12  y  +  28  2-168  =  0.    Ans. 

The  answer  should  be  checked  by  testing  whether  the  coordinates  of  Pj 
satisfy  the  answer. 

2.  To  find  the  equation  of  a  plane  passing  through  three  points,  sub- 
stitute for  X,  y,  and  z  in  (1)  the  coordinates  of  each  of  the  three  points. 
Then  three  equations  involving -4,  B,  C,  and  D  will  be  obtained,  which 
may  be  solved  for  three  o? these  coefficients  in  terms  of  the  fourth. 

It  is  convenient  to  write  down  the  equation  of  a  plane  passing  through 
three  given  points  (x^,  ?/j,  2j),  (x,,  y^,  z^),  {x^,  y^,  Zg)  in  the  form  of  a  deter- 
minant.  This  is 


(5) 


X 

y      z 

1 

X, 

y^  ^1 

1 

^2 

2/2       ^2 

1 

Xg 

Vz     Z3 

1 

=  0. 


In  fact,  when  (5)  is  expanded  in  terms  of  the  elements  of  the  first  row, 
an  equation  of  the  first  degree  in  x,  y,  and  z  results.  Hence  (5)  is  the 
equation  of  a  plane.  Further,  (5)  is  satisfied  when  the  coordinates  of 
any  one  of  the  three  given  points  are  substituted  for  x,  y,  and  2,  since 
then  two  rows  become  identical.  Hence  the  plane  (5)  passes  through  the 
given  points. 

The  equation  (5)  may  be  used  also  to  determine  whether  four  given 
points  lie  in  a  plane. 

If  we  write  (5),  when  expanded,  in  the  form 
Ax  +  By  -\-  Cz  +  D-  0, 
then  the  coefficients  are  the  determinants  of  the  third  order. 


A  = 


Vx    H    1 

X,     z,     1 

X,    y,    1 

Vi    z^    1 

,    B=- 

X2      2,      1 

,      C  = 

X.,    2/2    1 

Vz    23     1 

•*3      ~3      ^ 

■^3      Vz      1 

D=- 


Vi 
Vi 
Vz 


PROBLEMS 

Check  the  answer  in  each  of  the  following : 

1.  Find  the  equation  of  the  plane  which  passes  through  the  points 
(2,  3,  0),  {- 2,  -  3,  4),  and  (0,  6,  0).  Am.    3x-f22/-F62-r2  =  0. 

2.  Find  the  equation  of  the  plane  which  passes  through  the  points 
(1,  1,  -  1),  (-  2,  -  2,  2),  and  (1,  -  1,  2).  Ans.   x  -  3  //  -  2  2  =  0. 

3.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(3,  —  3,  2)  and  is  parallel  to  the  plane  3x  —  ?/-l-2  —  6  =  0. 

Ans.    3x  —  ?/  +  2  —  14  =  0. 


268  NEW  ANALYTIC   GEOMETRY 

4.  Find  the  equation  of  the  plane  which  passes  through  the  points 
(0,  3,  0)  and  (4,  0,  0)  and  is  perpendicular  to  the  plane  4x  —  6y  —  z=12. 

Ans.   3a;  +  4?/-12z-12  =  0. 

6.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(0,  0,  4)  and  is  perpendicular  to  each  of  the  planes  2x  —  Sy  =  b  and 
x-42  =  3.  Ans.    12x  +  8?/  +  3z-12  =  0. 

6.  Find  the  equation  of  the  plane  whose  intercepts  on  the  axes  are 
3,  5,  and  4.  Aiis.    20  x  +  12  2/  +  15  z  -  60  =  0. 

7.  Find  the  equation  of  the  plane  whicl*  passes  through  the  point 
(2,  —  1,  6)  and  is  parallel  to  the  plane  x  —  2y  —  3z  +  4  =  0. 

Ans.    x  —  2y-Sz  +  U  =  0. 

8.  Find  the  equation  of- the  plane  which  passes  through  the  points 
(2,  —  1,  6)  and  (1,  —  2,  4)  and  is  perpendicular  to  the  plane  x  —  2y  — 
22  +  9  =  0.  Ans.    2x  +  4?/-32  +  18  =  0. 

9.  Find  the  equation  of  the  plane  whose  intercepts  are  —  1,-1,  and  4. 

Ans.    4x  +  4?/  — 2  +  4  =  0. 

10.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(4,  —  2,  0)  and  is  perpendicular  to  the  planes  x  +  y  —  z  —  0  and  2  x  — 
4  ?/  +  2  =  5.  Ans.   X  +  y  +  2  z  —  2  =  0. 

11.  Show  that  the  four  points  (2,  -  3,  4),  (1,  0,  2),  (2,  -  1,  2),  and 
(1,  —  1,  3)  lie  in  a  plane. 

12.  Show  that  the  four  points  (1,  0,  -  1),  (3,  4,  -  3),  (8,  -  2,  G),  and 
(2,  2,  —  2)  lie  in  a  plane. 

13.  Find  the  equation  of  the  plane  which  is  perpendicular  to  the  line 
joining  (3,  4,  —  1)  and  (5,  2,  7)  at  its  middle  point. 

Ans.   X  —  ?/  +  42  —  13  =  0. 

14.  Find  the  equations  of  the  faces  of  the  tetrahedron  whose  vertices 
are  the  points  (0,  3,  1),  (2,  -  7,  1),  (0,  5,  -  4),  and  (2,  0,  1). 

A7W.    25x  +  5?/  +  22  =  17,  5x-22  =  8,  2  =  1,  15x  +  10?/  +  4z  =  34. 

15.  The  equations  of  three  faces  of  a  parallelepiped  are  x  —  4  ?/  =  3, 
2x  —  ?/  +  2  =  3,  and  Sx  +  y  —  2z  —  0,  and  one  vertex  is  the  point 
(3,  7,  —  2).    What  are  the  equations  of  the  other  three  faces  ? 

Ans.   X  —  4?/  +  25  =  0,  2x  —  ?/  +  2  +  3  =  0,  3x  +  ?/  —  22  =  20. 

16.  Find  the  equation  of  the  plane  whose  intercepts  are  o,  b,  c. 

Ans.  ?  +  ^  +  ?  =  l. 
a      b      c 

17.  What  are  the  equations  of  the  traces  of  the  plane  in  Problem  16? 
How  might  these  equations  have  been  anticipated  from  plane  analytic 
geometry  ? 


THE  PLANE  269 

18.  Find  the  equation  of  the  plane  which  passes  through  the  point 
P\  (Xp  2/1,  Zj)  and  is  parallel  to  the  plane  A^x  -}-  B^y  +  C^z  +  i*^  =  0. 

Aiis.    A^  (X  -  x^)  +  L',  {y  -  y^)  +  C\  {z  —  z^)  =  0. 

19.  Find  the  equation  of  the  plane  which  passes  through  the  origin  and 
Pj(Xj,  ^j,  z■^)  and  is  perpendicular  to  the  plane  J^jX  +  B^y  +  C-^z  +  Z>j  =  0. 

Ylns.    {B^z■^—  C^y^)x+  {C-^x^  —  A-^z^)y -{■  {A^y^—  B^x^)z  =  0. 

103.  The  equation  of  a  plane  in  terms  of  its  intercepts. 
Theorem.    If  a,  b,  and  c  are  respectively  the  intercejjfs  of  a  plane 

on  the  axes  of  X,  Y,  and  Z,  then  the  equation  of  the  jJ lane  is 

(III)  ^  +  |  +  _  =  1. 

^       -^  a       b       c 

Proof    Let  the  equation  of  the  required  plane  be 
(1)  Ax  +  Bij  -\-Cz-\-D=  0. 

•  Then  we  know  three  points  in  the  plane,  namely 

(./,  0,  0),     (0,^^,0),     (0,0,  f). 
These  coordinates  must  satisfy  (1).    Hence 

Aa  +  ])  =  0,     Bb-\-D  =  0,      Cc  +  D  =  0. 

Whence         A  = 

Substituting  in  (1),  dividing  by  —  D,  and  transposing,  we 
obtain  (III).  Q.  e.  d. 

104.  The  perpendicular  distance  from  a  plane  to  a  point.  The 
positive  direction  on  any  line  perpendicular  to  a  plane  is 
assumed  to  agree  with  that  on  the  line  drawn  through  the  ori- 
gin perpendicular  to  the  plane  (Art.  100).  Hence  the  distance 
fro7n  a  plane  to  the  point  P^  is  positive  or  negative  according 
as  Pj  and  the  origin  are  on  opposite  sides  of  the  plane  or  not. 

If  the  plane  passes  through  the  origin,  the  sign  of  tiie  distance  from 
the  plane  to  P,  must  be  determined  by  the  conventions  for  the  special 
cases  in  Art.  100. 


D 

D 

D 

— 5 

B=—  -, 

C  = 



a 

b 

c 

270 


NEW  ANALYTIC   GEOMETRY 


We  now  solve  the  problem  :  Given  the  equation  of  a  plane 
and  a  point,  to  find  the  perpendicular  distance  from  the  plane 
to  the  point. 

Solution.  Let  the  point  be  P^  (x^,  y^,  z^  and  assume  that  the 
equation  of  the  given  plane  is  in  the  normal  form 

(1)  ajcosar+z/cos^+^cosy— ^  =  0. 

Let  d  equal  the  required  distance. 

Dravsr  OP^.  Projecting  OP^  on 
ON,  we  evidently  get  j'j  +  d. 

Projecting  OE,  EF,  and  FP^  on 
ON,  we  get  respectively  x^  cos  a, 
y^  cos  /?,  and  z^  cos  y. 

Then,  by  the  second  theorem  of 
projection,  Y>( 

p  -\-  d  =  x^  cos  a  -\-  ij^  cos  fi  -\-  z^  cos  y. 
.•.  d  =  x^  cos  a  +  ^/j  cos  ji  -\-  z^  cos  y  —  p. 

Hence  the  perpendicular  distance  d  is  the  number  obtained 
by  substituting  the  coordinates  of  the  given  point  for  x,  y,  and 
z  in  the  left-hand  member  of  (1).    Whence  the 

Rule  to  find  the  perpendicular  distance  d  from  a  given  plane 
to  a  given  pjoint. 

Reduce  the  equation  of  the  plane  to  the  normal  form.  Place 
d  equal  to  the  left-hand  memher  of  this  equation. 

Substitute  the  coordinates  of  the  given  point  for  x,  y,  and  z. 
The  result  is  the  required  distance. 

For  example  :  To  find  the  perpendicular  distance  from  the  plane 
2a-  +  ?/  —  2z4-8  =  0  to  the  point  (—  1,  2,  3).  Dividing  the  equation 
by  —  3,  we  have 


d  = 


_  2x  +  J/-22  +  8  _  2(-l)+  2-2(3)  +  8 


:—  t-    Ans. 


-3  -3 

Hence  the  given  point  is  on  the  same  side  of  the  plane  as  the  origin. 


THE  PLANE  271 

The  rule  gives  for  the  perpendicular  distance  d  from  the  plane 
Ax  +  By  +  Cz  -{-  D  =  ^ 
to  the  point  (a^^  y^,  z^  the  result 
(2)  ^_^.,  +  %,+  C.,+£^ 

^  ±  V^2  ^_  ij2   _f_    (^,2 

the  sign  of  the  radical  being  determined  as  above  (Art.  101). 

105.  The  angle  between  two  planes.  The  plane  angle  of  one 
pair  of  dihedral  angles  formed  by  two  intersecting  planes  is  evi- 
dently equal  to  the  angle  between  the  positive  directions  of  the 
perpendiculars  to  the  planes.  That  angle  is  called  the  angle 
between  the  planes. 

Theorem.    The  angle  6  hetiveen  the  two  ■planes 

A^x  +  J5j//  +  C\z  +  Z)j  =  0  and  A^x  +  B,^j  +  C^z  +  D^=  Q  is 

given  by 

(IV)       cos^=  A,A,  +  B,B,  +  C,C, 


±^Al  +  Bi  +  C,2  X  ±  V>1 1  +  Bi  +  Ci 
the  signs  of  the  radicals  being  chosen  as  in  Art.  101. 

Proof.    By  definition  the  angle  Q  between  the  planes  is  the 
angle  between  their  normals. 

The  direction  cosines  of  the  normals  to  the  planes  are 


cos  CTi 

= 

^1 

± 

V.42 

+  BI 

+  ci 

COS/81 

= 

B, 

± 

VI7 

+  B^ 

+  c\^ 

cos  V, 

c\ 

COS  a.2 

= 

A. 

± 

^Ji 

+  BI 

+  CI 

cos  13.2 

= 

B, 

± 

-Jai 

-\-B.I 

+  c| 

cos  v„ 

c. 

±  V.4 -^  +  Bf  +  C^  '       ±  V^l  +  Bl  +  C| 

By  (V),  Art.  90,  we  have 

cos  Q  —  cos  a^  cos  a^  +  cos  ^j  cos  ^^  -\-  cos  y^  cos  y„. 

Substituting   the   values    of   the    direction    cosines    of   the 
normals,  we  obtain  (IV).  q.e.d. 


272  NEW  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Find  the  distance  from  the  plane 

(a)  6x  —  3?/  +  2z  -  10  =  0  to  the  point  (4,  2,  10).  Ans.  4t. 

(b)  X  +  2?/  — 2z  — 12  =  0  to  the  point(l,  -  2,  3).  Ans.  —7. 

(c)  4x  +  3?/ +  122  +  0  =  0  to  tlie  point(9,  -1,  0)  Ans.  —  3.__ 

(d)  2x- 5?/  + 3z  -  4  =  0  to  the  point(.- 2,  1,  7).  .     Ans.  i'VV38. 

2.  Do  the  origin  and  tlie  point  (3,  5,  —  2)  lie  on  the  same  side  of  the 
plane  7x  —  y  —  3z  +  Q  =  0?  Ans.   Yes. 

3.  Does  the  point  (1,  6,  0)  lie  on  the  same  side  of  the  plane 
X  +  2y  —  3z  =  6  as   the   origin  ? 

4.  Find  the  length  of  the  altitude  which  is  drawn  from  the  first  vertex 
of  the  tetrahedron  whose  vertices  are  (0,  3, 1),  (2,  —  7, 1),  (0,  5,  —  4),  and 
(2,  0,  1).  Ans.    i^  V29. 

5.  Find  the  volume  of  the  tetrahedron  formed  by  the  point  (1,  2,  1) 
and  the  points  where  the  plane  3x+4?/  +  22  —  12  =  0  intersects  the 
coordinate  axes. 

6.  Find  the  volumes  of  the  tetrahedrons  having  the  following  vertices  : 

(a)  (3,  4,  0),  (4,  -  1,  0),  (1,  2,  0),  (6,  -  1,  4).  Ans.    8. 

(b)  (0,  0,  4),  (3,  0,  0),  (0,  2,  0),  (7,  7,  3). 

(c)  (4,  0,  0),  (0,  4,  0),  (0,  0,  4),  (7,  3,  2). 

(d)  (3,  0,  0),  (0,  -  2,  0),  (0,  0,  -  1),  (3,  -  1,  -  1).  Ans.    |. 

(e)  (1,  0,  0),  (0,  1,  0),  (0,  0,  -  2),  (4,  -  1,  .3). 

(f )  (3,  0,  0),  (0,  5,  0),  (0,  0,  -  1),  (3,  -  4,  0). 

7.  Find  the  angles  between  the  following  pairs  of  planes  : 

(a)  2x  + ij —  •2z  —  9  =  0,x  —  2y +  2z  =  0.  Ans.    cos-i(— |). 

(b)  X  +  2/  -  4z  =  0,  3  ?/  -  3  2  +  7  =  0. 

(c)  4x  + 2?/  +  4z-7  =  0,  3x- 4?/ =  0. 

(d)  2x-2/  +  z  =  7,  x  +  ?/  +  2z  =  11. 

(e)  3X-22/  + 6z  =  0,  x  + 2?/-2z  + 5  =  0. 

(f)  x  +  5?/-3z  + 8  =  0,  2x-3i/  +  z-5  =  0. 

8.  Show  that  the  angle  given  by  (V)  is  that  angle  formed  by  the  planes 
•which  does  not  contain  the  origin. 

9.  Find  the  vertex  and  the  dihedral  angles  of  that  trihedral  angle 
formed  by  the  planes  x  +  7/  +  z  =  2,  x  —  y  —  2z  =  4,  and  2x+?/  —  z  =  2  in 
which  the  origin  lies.     .         ,,         ,    „,  ,  1    /-  27r 


Ans. 

cos-i(- 

Ans. 

cos-i 1 

Ans. 

cos-i(- 

Ans. 

TT 

Ans.  (4,  -  4,  2),  cos-i  -  V2,  — ,  cos-i  (-  -  V2'\ 
3  3  \     3       / 


the'  plane  273 

10.  Find  the  equation  of  the  plane  which  passes  through  the  points 

27r 
(0,  —  1, 0)  and  (0,  0,  —  1)  and  which  makes  an  angle  of  —  with  the  plane 

Ans.   ±  V6x  +  ?/  -f  z  +  1  =  0. 

11.  Find  the  locus  of  points  which  are  equally  distant  from  the  jjlanes 
2x  —  2/  —  22-3  =  0  and  6x  —  32/  +  2z  +  4  =  0. 

Am.    32x-162/-8z- 9  =  0. 

12.  Find  the  locus  of  a  point  which  is  three  times  as  far  from  the  plane 
3 J  _  6 ?/  —  2 z  =  0  as  from  the  plane  2x— ?/  +  2z  =  9. 

Ans.    17x-13y  +  122- 63  =  0. 

13.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
plane  x  +  2/+z  —  l=Ois  equal  to  its  distance  from  the  origin. 

14.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
plane  x  +  y  =  1  equals  its  distance  from  the  Z-axis. 

Ans.  (X  -  j^)2  +  2  (X  +  ?/)  -  1  =  0. 

15.  Find  the  equation  of  the  locus  of  a  point,  the  sum  of  the  squares  of 
whose  distances  from  the  planes  x  +  y  —  z— 1  =  0  and  x  +  ?/  +  z  +  l  =  0 
is  equal  to  unity.  Ans.   2(x  +  ?/)2 +  2  z(z  +  2)  —  1  =  0. 

106.  Systems  of  planes.  The  equation  of  a  plane  which  sat- 
isfies two  conditions  will,  in  general,  contain  an  arbitrary  con- 
stant, for  it  takes  three  conditions  to  determine  a  plane.  Such 
an  equation  therefore  represents  a  systevi  of  planes. 

Systems  of  planes  are  used  to  find  the  equation  of  a  plane 
satisfying  three  conditions  in  the  same  manner  that  systems 
of  lines  are  used  to  find  the  equation  of  a  line  (Art.  36). 

Three  important  systems  of  planes  are  the  following : 

The  system  of  planes  parallel  to  a  rj'iv  en  plane 

Ax  +  ny-^  Cr.  -f  X>  =  0 
is  represented  by 

(V)  Ax  +  By-irCz  +  k  =  Q, 

where  k  is  an  arbitrary  constant. 

The  plane  (V)  is  obviously  parallel  to  the  given  plane  (Corol- 
lary I,  Art.  101). 


274  NEW  ANALYTIC  GEOMETRY 

The  system  of  planes  passing  through  the  line  of  intersection 
of  two  given  planes 

A^x  +  B^j  +  C^z  +  i),  =  0,     A^x  +  i3,3/  +  ( './  +  il.  =  0 
is  represented  by 

(VI)  A^x  +  B^y  +  Qz  +  Z?i  +  ft  {A^x  +  B^y  +  Qz  +  ZJ^)  =  0, 

where  k  is  a>i  arbitrary  constant. 

Clearly,  the  coordinates  of  any  point  on  the  lifte  of  intersec- 
tion will  satisfy  the  equations  of  both  of  the  given  planes,  and 
hence  will  satisfy  (VI)  also. 

The  equation  of  a  system  of  planes  which  satisfy  a  single 
condition  must  contain  two  arbitrary  constants.  One  of  the 
most  important  systems  of  this  sort  is  the  following : 

The  system  of  planes  passing  throngli  n  gir  en  point  I\(x-^,  y^,  z^ 
is  represented  by 

(VII)  ^(x-jrO  +  5(i/-i/0+C(z-zO  =  0. 

Equation  (VII)  is  the  equation  of  a  plane  which  passes 
through  P^,  for  the  coordinates  of  P^  obviously  satisfy  it. 
Again,  if  any  plane  whose  equation  is 

Ax  +  7>'//  +  Cz  +  /)  =  0 

passes  through  7^^,  then 

Subtracting,  we  get  (VII).  Hence  (VII)  represents  all 
planes  passing  through  P^. 

Equation  (VII)  contains  two  arbitrary-  constants,  namely,  the 
ratio  of  any  two  coefficients  to  the  third. 

In  the  following  problems  write  down  the  equation  of  the 
appropriate  system  of  planes  and  then  determine  the  unknown 
parameters  from  the  remaining  data. 


THE  PLANE  275 

PROBLEMS 

1.  Determine  the  value  of  k  such  that  the  plane  x  +  fcy  —  2z  —  9  =  0 

shall 

(a)  pass  through  the  point  (5,  —  4,  —  6).  Ans.  2. 

(b)  be  parallel  to  the  plane  &x  —  2y  —  Viz  =  1.  Aim.  —  \. 

(c)  be  perpendicular  to  the  plane  2j;  —  4?/  +  2  =  3.  Ans.  0. 

(d)  be  3  units  from  the  origin.  Ans.  ±  2. 

IT 

(e)  make  an  angle  of  —  with  the  plane  2a;  —  2t/  +  z  =  0. 

^  Ans.    -  3  V35. 


2.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(3,  2,-1)  and  is  parallel  to  the  plane  7  x  —  ?/  +  z  =  14. 

Ans.    7x  — y  +  z  — 18  =  0. 

3.  Find  the  equation  of  the  plane  which  passes  through  the  inter- 
section of  the  planes  2x-|-?/— 4  =  0  and  j/  +  2  2  =  0,  and  which 
(a)  passes  through  the  point  (2,  —  1,  1);  (b)  is  perpendicular  to  the  plane 
3x-l-2?/-3z  =  6. 

Ans.    {a)x  +  2/  +  z-2  =  0;  (b)2x  +  3?/  +  4z-4  =  0. 

4.  Find  the  ecjuations  of  the  planes  which  bisect  the  angles  formed 
by  the  planes 

(a)  2  X  —  ?/  +  2  z  ^  0  and  x  +  2  y  —  2  z  =  6. 

Ans.    3x  +  ?/-6  =  0,  X-3  2/  +  4Z+6  =0. 

(b)  6x  —  22/-3z  =  0  and  4  x  +  3  ?/  -  13  z  =  10. 

5.  Find  the  equations  of  the  planes  passing  through  the  line  of  inter- 
section of  the  planes  2x-f-i/  —  z  =  4  and  x— ?/-i-2z  =  0  which  are  per- 
pendicular to  the  coordinate  planes. 

Ans.    5x4-2/  =  8,  3x4-2  =  4,  3y  — 5z  =  4. 

6.  Find  the  equation  of  a  plane  parallel  to  the  plane  6x— 32/  + 2z-f- 21  =  0 
and  tangent  to  a  sphere  of  unit  radius  whose  center  is  the  origin. 

7.  Find  the  equationof  aplane  parallel  to theplane6x— 2?/— 3z-|-35=0 
and  such  that  the  point  (0,  —2,-1)  lies  midway  between  the  two  planes. 

8.  Find  the  equation  of  a  plane  through  the  point  (2,  —  3,  0),  and 
having  the  aame  trace  on  the  ^Z-plane  as  the  plane  x  —  3?/  +  7z  —  2  =  0. 

9.  Find  the  equationof  aplane  parallel  to  the  plane  2x-l-2/-|-2z  +  5  =  0, 
and  forming  a  tetrahedron  of  unit  volume  with  the  three  coordinate 
planes. 

10.  Find  the  equation  of  aplane  parallel  to  the  plane  5x-l-3  j/-|-  z  — 7=0 
if  the  sum  of  its  intercepts  is  23. 


270  NEW  ANALYTIC   GEOMETRY 

11.  Find  the  equation  of  a  plane  parallel  to  the  plane  2x  +  6y + 
8^  —  8  =  0,  upon  which  the  area  intercepted  by  the  coordinate  planes 
in  the  first  octant  is  |.  Ans.    2x  +  6?/  +  3z  —  3  =  0. 

12.  Find  the  equation  of  a  plane  parallel  to  the  plane  2x  +  y  + 
2z  —  5  =  0  and  such  that  the  entire  surface  of  the  tetrahedron  which  it 
forms  with  the  coordinate  planes  is  unity.      Ans.    2x  +  y  +  2z±l=0. 

13.  Find  the  equation  of  a  plane  having  the  trace  x  +  Sy  —  2  =  0  and 
forming  a  tetrahedron  of  volume  |  with  the  coordinate  planes. 

Ans.   3x  +  9y  +  z  —  6  -0. 

14.  Find  the  equation  of  a  plane  passing  through  the  intersection  of 
the  two  planes  6x-\-2y+Sz  —  G  =  0  and  x  +  y  +  z  —  1  —  0  and  forming 
a  tetrahedron  of  unit  volume  with  the  coordinate  planes. 

Ans.    12x-8y  -3z-12  =  0. 

15.  A  point  moves  so  that  the  volume  of  the  tetrahedron  which  it 
forms  with  the  three  points  (2,  0,  0),  (0,  6,  0),  and  (0,  0,  4)  is  always 
equal  to  2.   Find  the  equation  of  its  locus. 

16.  A  point  moves  so  that  the  sum  of  its  distances  from  the  three 
coordinate  planes  is  unity.  Determine  the  equation  of  the  locus  of  a 
second  point  which  bisects  the  line  joining  the  first  with  the  origin. 

17.  Find  the  equation  of  the  plane  passing  through  the  intersection  of 
the  planes  A^x  +  B^y  +  C^z  +  1)^-0  and  A.,x  +  B„y  +  C^z  +  D^  =  0 
which  passes  through  the  origin. 

Ans.    (A^D^  -  A^D^)x  +{B^D,  -  B.-,D^)  y  +  {C^D.^-  C.^D^)z  =  0. 

18.  Find  the  equations  of  the  planes  which  bisect  the  angles  formed 
by  the  planes  A^x  +  B^y  +  C^z  +  2>j  =  0  and  A„x  +  B^y  +  C^z  +  Dg  =  0. 

.         A,x  +  B^y  +  C^z  +  Di       ,  A^x  +  B-^y  +  C^z  +  B^ 
Ans.    — i —      ^  '      — -  =  +  — - —  ^       — -• 

19.  Find  the  equations  of  the  planes  passing  through  the  intersection 
of  the  planes  A^x  +  B^y  +  CjZ  +  D,  =  0  and  A^x  +  B^y  +  CgZ  +  ^2  =  *^ 
which  are  perpendicular  to  the  coordinate  planes. 

Ans.  {A^B^  -  A„B^)y-  {C^A^  -  C,A^)z  +  A^D^  -  A^D^  =  0, 
(^^^2  -  ^2^i)a;  -{B,C2  -  B.^C^)z-  {B^D^  -  B^l),)  =  0, 
(C'1^2  -  ^2^1)0;  -  (B1C2  -  BiC^)y  +  C^D^  -  C^I)^  =  0. 


CHAPTER  XVI 

THE  STRAIGHT  LINE  IN  SPACE 

107.  General  equations  of  the  straight  line.  A  straight  line 
may  be  regarded  as  the  intersection  of  any  two  planes  which 
pass  through  it.  The  equations  of  the  planes  regarded  as 
simultaneous  are  the  equations  of  the  line  of  intersection, 
and  hence  the  . 

Theorem.  The  equations  of  a  straight  line  are  of  the  first 
degree  in  x,  ?/,  and  z. 

Conversely,  the  locus  of  two  equations  of  the  first  degree  is  a 
straight  line  unless  the  planes  which  are  the  loci  of  the  separate 
equations  are  parallel.    Hence  w^e  have  the 

Theorem.    The  locus  of  two  equations  of  the  first  degree, 

is   «   straight   line   unless   the   coefficients  of  x,   y,  and  z   are 
proportional. 

To  plot  a  straight  line  we  need  to  know  only  the  coordinates 
of  two  points  on  the  line.  The  easiest  points  to  obtain  are 
usually  those  lying  in  the  coordinate  planes,  which  we  get  by 
setting  one  of  the  variables  equal  to  zero  and  solving  for  the 
other  two,  as  in  the  following  example. 

The  direction  of  a  line  is  known  when  its  direction  cosines 
are  known.  The  method  of  obtaining  these  will  now  be 
illustrated. 

277 


278  NEW  ANALYTIC  GEOMETRY 

EXAMPLES 

1.  Find  the  direction  cosines  of  the  line  whose  equations  are 
(1)  Sx  +  2y-z-l  =  0,     2x-2/  +  2z-3  =  0. 

Solution.  Let  us  find  the  point  where  the  line  pierces  the  A"F-plane. 
To  do  this,  let  z  =  0  in  both  eijuations.  Then  solving  the  resulting 
equations  3x  +  2y  —  1  =  0  and  2x  —  ?/  —  3  —  0  for  x  and  ?/,  we  find  the 
requii-ed  point  is  (1,  —  1,  0).  Similarly,  putting  y  =  0,  the  point  on  the 
line  in  the  ZJT-plane  is  (f,  0,  |). 

Hence  A  (1,  —  1,  0)  and  B  (|,  0,  ^)  are  two  points  on  the  line. 

Let  the  required  direction  cosines  of  AB  be  cos  a,  cos/3,  and  cos 7. 
Then,  by  the  corollary  of  Art.  89, 

cos  a  _    cos  /3     _  cos  7  _ 

or,  reducing  (multiplying  the  denominators  by  8), 

cos  a  _  cos  /3  _  cos  7 
^  '  3      ~  -8  ^   -  7  ■ 

The  direction  cosines  may  now  be  found  as  usual  (Art.  88). 
A  second  method  is  the  following  : 

,  ,^    ,  cos  a      cosS      cos  7 

(4)  Assume  = = 

^  '  a  b  c 

The  coefficients  3,  2,  and  —  1  in  the  first  plane  of  (1)  are  proportional 
to  the  direction  cosines  of  a  perpendicular  to  that  plane.  The  required 
line  lies  in  this  plane.    Hence  (corollary.  Art.  90) 

(5)  3a  +  26-c  =  0. 

For  the  same  reason,  using  the  second  plane  in  (1), 

(6)  2a-b  +  2c  =  0. 

Solving  (5)  and  (6)  for  the  ratios  of  a,  6,  and  c,  the  result  is 
8a=-36,     7a  =  -3c. 

(7)  ...^  =  -^  =  ^. 
^ '  3-8-7 

Combining  (7)  and  (4),  we  have  the  previous  result  (3). 

2.  Find  the  direction  cosines  of  the  line  (I). 

Solution.    The  direction  cosines  cos  a,  cos  ;3,  cos  7  must  satisfy 
Aj^cof^a  +  B^  cos  ^  +  C^  cos  7  =  0,      A^  cos  a  +  B^cos^  +  C^  cos  7  =  0, 
reasoning  as  in  Ex.  1. 


THE  STRAIGHT  LINE  IN  SPACE  279 

Solving  these  equations  for  the  ratios,  we  have  the 
Theorem.   If  a,  /3,  and  7  are  the  direction  angles  of  the  line  (J),  then 
cos  a        _  cos  P  _  cos  -y 

B1C2—B2C1  ~  C1A2—C2A1  ~  A1B2—A2B1 ' 

The  denominators  are  readily  remembered  as  the  three  determinants 


of  the  second  order 


B,     C, 
B„     C„ 


C,     A^\       \A,     B, 
Co     A^\       \  At,     Bg 


'2       ^2 

formed  from  the  coefficients  of  x,  ?/,  and  z  in  (I). 

PROBLEMS 

1.  Find  the  points  in  which  the  following  lines  pierce  the  coordinate 
planes,  and  construct  the  lines : 

(a)  2x  +  2/-z  =  2,  X-2/  +  22-4.  (c)  a;  +  2?/ =  8,  2x- 4?/ =  7. 

(b)  4x  +  3y-6z  =  12,  4x-32/  =  2.      (d)  ?/ +  z  =  4,  X-2/  +  22  ==  10. 

2.  Find  the  direction  cosines  of  the  following  lines : 

(a)  2X-2/  + 2z  =  0,  x  + 22/-2Z  =  4.  _ 

Ans.    i^^sVeS,  Tg'VV^,  Tt\V65. 

(b)  x  +  2/  +  z  =  5,  x-?/  +  z  =  3.  An&.    ±\  V2,  0,  T^  V2. 

(c)  3x  +  22/-z  =  4,x-2y-2z  =  5.  ^TW.    i^^Vs,  TiVs,  ±5*5 VB. 

(d)  x  +  ^-3z  =  6,2x-?/  +  3z  =  3.  ^ns._0,  ±t%V10,  ±i:^-\^. 

(e)  x  + ?/  =  6,  2x-3z  =  5.  Ans.    ±^35V22,  T3j\V22,  ±JjV22. 

(f)  2/4- 3z  =  4,3  2/- 5z  =  l.  ^«s.    ±1,_0,  0. 

(g)  2x-3y  +  2  =  0,  2x-32/-2z  =  6.    Ans.    ±^3 Vl3,  i^^g Vl3,  0. 
(h)  5x-142-7  =  0,  2x  +  7z  =  19.  Ans.   0,  ±1,  0. 

3.  Show  that  the  lines  of  the  following  pairs  are  parallel  and  construct 
the  lines : 

(a)  2  2/  +  z  =  0,  3  2/  -  4  z  =:  7  ;  and  5  ?/  -  2  z  =  8,  4  2/  +  H  z  =  44. 

(b)  x  +  22/-z=7,2/+z  — 2x=:6;and3x  +  6^— 3z  =  8,2x-2/-2  =  0. 

(c)  3x  +  z  =  4,  2/  +  2z  =  9;  and  6x  —  2/  =  7,  3  2/  +  6z  =  l. 

4.  Show  that  the  lines  of  the  following  pairs  meet  in  a  point  and  are 
perpendicular : 

(a)  x  +  22/  =  l,  2?/  —  z  =  l;  and  x  —  2/  =  l,  x  —  2z  =  3. 

(b)  4x  +  2/  —  32  +  24  =  0,  z  =  5;  and  x  +  ?/4-3  =  0,  x  +  2  =  0. 

(c)  3x  +  2/-z  =  l,2x— 2  =  2;  and2x  —  ?/  +  22  =  4,  x  — 2/+22  =  3. 


280 


NEW  ANALYTIC   GEOMETRY 


5.  Find  the  angles  between  the  following  lines,  assuming  that  they  are 
directed  upward,  or  in  front  of  the  ZX-plane : 

77" 

(a)  x  +  y  —  z~0,  y+z  =  0;  and  x  —  y  =  1,  x  —  3  y  +  z  =  0.  Ans.    -. 

o 

(b)  x  +  2y  +  2z  =  l,x  —  2z=zl;  and  4,x  +  3//-z  +  l=0,  2x  +  32/=0. 

Ans.    cos-i^^. 

(c)  X  —  2y+z  =  2,2y— z  =  l;  and  x  —  2y+z  =  2,  x  —  2y  +  2z  =  4. 

Ans.    cos-i^. 

6.  Find  the  equations  of  the  planes  through  the  line 

X  +  2/  —  z  =  0,  2  X  —  ?/  +  3  2  =  5, 

which  are  perpendicular  to  the  coordinate  planes. 

Ans.   3x  +  2z  =  5,  3r/— 5z  +  5  =  0,  5x  +  22/  =  5. 

7.  Show  analytically  that  the  intersections  of  the  planes  x  —  2y  —  z  =  3 
and  2x  —  4?/  —  2z  =  5  with  the  plane  x  +  y  —  3z  =  0  are  parallel  lines. 

8.  Verify  analytically  that  the  intersections  of  any  two  parallel  planes 
with  a  third  plane  are  parallel  lines. 

108.  The  projecting  planes  of  a  line.  The  three  planes  pass- 
ing through  a  given  line  and  perpendicular  to  the  coordinate 
planes  are  called  the  project- 
ing planes  of  the  line. 

If  the  line  is  perpendicular  to 
one  of  the  coordinate  planes,  any 
plane  containing  the  line  is  per- 
pendicular to  that  plane.  In  thi.-^ 
case  we  speak  of  but  two  project- 
ing planes,  namely,  those  drawn 
through  the  line  perpendicular 
to  the  other  coordinate  planes. 

If  the  line  is  i^arallel  to  one 
of  the  coordinate  planes,  two  of 
the  projecting  planes  coincide. 

By  (VI),  Art.  106,  the  equation  of  any  plane  through  the  line 

(1)  3.'K-f22/-^-l  =  0,     2x  -y  +  2z-S  =  0 

has  the  form 


3  .^  +  2  ?/  -  «  -  1  -f  /o  (2  .T  -  y  +  2  ,-  -  3)  =  0. 


THE  STRAIGHT  LINE  IN  SPACE  281 

Multiplying  out  and  collecting  terms, 

(2)  (3  +  2k)x+(2-k)7/+{-l-\-2k)z-l-Sk  =  0. 

This  plane  will  be  perpendicular  to  the  XF-plane  when  the 
coefficient  of  z  equals  zero,  that  is,  if  A;  =  ^.  Writing  this  value 
of  k  in  (2)  and  reducing, 

(3)  4a-  +  |y-f  =  0,     0TSx  +  3y  -5  =  0. 

This  is  therefore  the  equation  of  the  projecting  plane  of  the 
line  (1)  on  XY,  that  is,  of  the  plane  ABA^D^  of  the  iigure. 

Now  equation  (3)  is  simply  the  result  obtained  by  cUmlnotlng 
zfrom  the  equations  (1);  namely,  we  multiply  the  lirst  of  equar 
tions  (1)  by  2  and  add  it  to  the  second.    Hence  the  result : 

To  find  the  ejnatlons  of  the  prqjectlng  planes  of  (i  line.,  elim- 
inate X,  y,  and  z  In  turn  from  the  gioen  equations. 

Thus,  to  finish  the  example  begun,  eliminating  //  from  (1), 
we  find  7  ic  +  3  ;s  —  7  =  0  for  the  projecting  plane  on  XZ. 
Eliminating  r,  we  get  7?/  —  8«4-7  =  0  for  the  equation  of 
the  projecting  plane  on  YZ. 

Special  forms  of  the  projecting  planes  will  indicate  special 
positions  of  the  line  relative  to  the  coordinate  planes.  These 
cases  should  be  noted  in  the  following  problems. 

PROBLEMS 

1.  Find  the  equations  of  the  projecting  planes  of  the  following  lines; 

(a)  2  X  +  ?/  —  z  =  0,  x  —  ?/  +  2  z  =  3. 

Arts.   6a;  +  2/  =  8,  3x  +  z  =  8,  3(/-5z  +  (5  =  0, 

(b)  a;  +  7/  +  z  =  0,  z  —  ?/  —  2z  =  2. 

Ans.    3x  +  2/ =  14,  2a;  — z  =  8,  2?/ +  3z  =  4. 

(c)  2x  +  y  — z  =  1,  X  — 2/ +  z  =  2. 

Ans.    Line  parallel  to  YZ.   x  =  l,  ?/  —  z  +  l  =  0. 

(d)  x  +  y-4z  =  l,2x  +  2?/  +  z  =  0. 

Ans.   Line  parallel  to  XY.    9x  +  Oy  =  1,  9z  +  2  -  0. 

(e)  2y  +  3z  =  G,  2?/-3z  =  18. 

Ans.   Line  parallel  to  OX.   ?/  =  0,  z  =—  2. 

(f)  2x-?/+ z  =  0,  4x  +  3;/  + 2z  =  0.    Ans.   6?/ =  G,  lOx  +  5z  =  0. 

(g)  X  +  z  =  1,  X  —  z  =  3.  Ans.   x  =  2.  z  =  —  1. 


282  NEW  ANALYTIC  GEOMETRY 

2.  Reduce  the  equations  of  the  following  lines  to  the  given  answers  and 
construct  the  lines : 

(a)x  +  2/  — 22  =  0,  X  — y  +  2  =  4.  Ans.  x  =  lz  +  2,  y  =^^z  —  2. 

{h)  x  +  2y  —  z  =  2,  2x  + 4y +  2z  =  5.  Ans.  z  =  \,  y  =-  Ix  +  I. 

(c)  X  —  2y  +  z  =  'i,  x  +  2y  —  z  =  Q.  Ans.  x  =  5,  y  =  iz  +  l. 

(d)  X  +  3z  =  6,  2x  +  6z  =  8.  Ans.  z  =  4,  x  =-  6. 

(e)  X  +  22/ —  2z  =  2,  2x  +  ?/  — 4z  =  1.  Ans.  x  =  2z,y  =  l. 
(i)  x  —  y  +  z  =  S,3x-3y  +  2z  =  6.  Ans.  z  =  3,  y  =  x. 

3.  Find  the  equations  of  the  line  passing  through  the  points  {—  2,  2,  1) 
and  (-  8,  5,  -  2).  Ans.   x  =  2z-4,  ?/=-z  +  3. 

4.  Find  the  equations  of  the  projection  of  the  line  x  =  z4-2,  y  =  2z  —  i 
upon  the  plane  x  +  y  —  z  =  0.  Ans.   x  =  ^ z  +  \^,  y  =  iz—  y . 

5.  Find  the  equations  of  the  projection  of  the  line  z  =  2,  y  =  x  —  2 
upon  the  plane  x  —  2y  —  3z  =  i.  Ans.   x=— 5z  +  4,  y  =—  4z. 

6.  Show  that  the  equations  of  a  line  may  be  written  in  one  of  the  forms 


( y  =  nix  +  a^        j  x  —  a,  Jx 

\z  =  nx  +b,        iz  =  my  +  b,       \y 


according  as  it  pierces  the  FZ-plane,  is  parallel  to  the  FZ-plane,  or  is 
pai-allel  to  the  Z-axis. 

7.  Show  that  the  condition  that  the  line  x  =  mz  +  a,  y  =  nz  +  b  should 

,     . .  ,  ,  /        ,,  .     a  —  a'        h  —  h' 

intersect  the  line  x  =  m  z  +  a .,  y  =  n  z  ■\-  b  is 


m       n  —  n 

109.  Various  forms  of  the  equations  of  a  straight  line. 

Theorem.  Parametric  form.  The  coordinates  of  any  point 
P  (x,  I/,  z)  on  the  line  through  a  given  jJolnt  Pi(x^,  y^,  z^)  whose 
direction  angles  are  a,  jS,  and  y  are  given  by 

(II)      x  =  Xj^-\-  pcosa,     y  =  y^-^  p  cos  p,     z  =  Zj^-\-  p  cosy, 
where  p  denotes  the  variable  directed  length  P^P- 

Proof.    The  projections  of  P^P  on  the  axes  are  respectively 

^  -  ^v     y  -  Vv     •'  -  ^r 
But,  by  the  first  theorem  of  projection,  these  are  also  equal  to 

p  cos  a,     p  cos  p,     p  cos  y. 
Hence 
X  —  x^  —  p  cos  <t.     11  —  //|  Tz^  p  cos  yS,     z  —  z^  =  p  cos  y. 


THE  STRAIGHT  LINE  IN  SPACE  283 

Solving  for  x,  y,  and  z,  we  obtain  (II).  Q.  e.l. 

Theorem.  Symmetric  form.  The  equations  of  the  line  passing 
through  the  point  P-^Qc^,  y^,  z^  tvhose  direction  angles  are  a,  fi, 
and  y  have  the  form 

(III)  x-x^^y-y^^z-Zj^ 

cos  a        cos^        cosy 

To  obtain  (III),  solve  each  of  the  equations  of  (II)  for  p  and 
equate  results. 

cos  a      cos  3      cos  v 
Corollary.    If =  — - —  = *• ;  then  the  symmetric  equa- 
tions of  the  line  may  be  ivritten  in  the  form 

(IV)  x-x^^y-y^^z-z^ 
^     ^  a  b  c 

Theorem.  Two-point  form.  The  equations  of  the  straight  line 
passing  through  P^{x^,  y^  z^  and  -P2(^2'  2^2'  ^2)  ^''"^ 

^    ^  x^-x^      y^-y^      z^-z^ 

Proof.  The  line  (III)  passes  through  P^.  If  it  also  passes 
through  P^,  then  the  coordinates  cc.,,  y,^,  and  z.^  may  be  substi- 
tuted for  X,  y,  and  z,  and  therefore 

^2  -  »'i  ^  2/2  -  Ih  ^  ^2-^1 
cos  a  cos  (5         cos  y 

Dividing  (III)  by  this  result,  we  obtain  (V).  q.e.d. 

Equations  (III)-(V)  each  involve  three  equations,  namely 
those  obtained  by  neglecting  in  turn  one  of  the  three  ratios. 
These  equations  are,  in  different  form,  the  equations  of  the 
projecting  planes,  since  one  variable  is  lacking  in  each.  Any 
two  of  the  three  equations  are  independent  and  may  be  used 
^s  the  equations  of  the  line,  but  all  three  are  usually  retained 
for  the  sake  of  their  symmetry.  In  (IV)  and  (V),  note  that  the 
denominators  are  numbers  proportional  to  the  direction  cosines 
of  the  line. 


284  NEW  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Find  the  equations  of  the  lines  which  pass  through  the  following 
p&irs  of  points,  reduce  them  to  the  given  answers,  and  construct  the  lines  : 

(a)  (3,  2,  -  1),  (2,  -  3,  4).  Ans.   x  =-  Iz  +  ^,  y  =- z  +  1. 

(b)  (1,  6,  3),  (3,  2,  3).*  Ans.  z  =  Z,y=-2x  +  8. 

.(c)  (1,  -  4,  2),  (8,  0,  3).  Ans.   x  =  2z  -  3,  y  =  4z  -  12. 

id)  (2,  -  2,  -  1),  (8,  1,  -  1).  Ans.   z=-l,y  =  3x-S. 

■(e)  (2,  3,  5),  (2,  -  7,  5).  Ans.   z  =  5,  x  =  2. 

2.  Show  that  the  two-point  form  of  the  equations  of  a  line  becomes 
5-  =  — ^  ,2  =  2.,  if  2.  =  Zo-    What  do  they  become  if  y.=y„? 

^i  -  ^\    y-2  -  V\ 

if  Xj  =  Xg  ? 

3.  What  do  the  two-point  equations  of  a  line  become  if  x^  =  x^  and 
y^=y^'>    if  ?/,  =  j/g  and  2^  =z,,  ?    if  z,  =  z^  and  x^  =  Xg  =* 

4.  Do  the  following  sets  of  points  lie  on  sti-aight  lines  ? 

(a)  (3,  2,  -  4),  (5,  4,  -  6),  and  (9,  8,  -  10).  Ans.    Yes. 

(b)  (3,  0,  1),  (0,  -  3,  2),  and  (G,  3,  0).  Ans.    Yes. 

(c)  (2,  5,  7),  (-  3,  8,  1),  and  (0,  0,  3).  Ans.    No. 

5.  Show  that  the  conditions  that  the  three  points  P^(Xj,  y■^,  Zj), 
^2  (•''2'  2/21  ^2)'  ^'^^^   ^s  (•''S'  Vzi  %)  should  lie   on  a  straight  line  are 

X2  -  Xj       y.-,  -  2/j      Zo  —  z, 

6.  Find  the  equations  of  the  line  passing  through  the  point  (2,  —  1,-3) 
whose  direction  cosines  are  proportional  to  3,  2,  and  7,  and  reduce  them 
to  the  given  answer.  Ans.   x  =  ^z+^f',y  =  fZ— }. 

7.  Find  the  equations  of  the  line  passing  through  the  point  (0,  —  3,  2) 

which  is  parallel  to  the  line  joining  the  points  (3,  4,  7)  and  (2,  7,  5). 

.         X      y+3      z-2 

Ans.    -  = = 

1        -3  2 

«o,         .,    ..,      1-        X  — 2      y  +  2      z       ,x  +  l      ?/  — 5      z  +  3 

8.  Show  that  the  lines  =      ^     =  -  and  — ■ —  = =  -- — 

are  parallel.  ""  ~  ~ 

X  —  I     u  —  6     2  —  3 
*  From  (V),  - — -  =  ^ — -  =  - — -  •   The  value  of  the  last  ratio  is  infinite  unless 

2-3=0.   If  2  -  3  =  0,  then  the  last  ratio  may  have  any  value  and  may  be  equal 

X  —  1     V  —  6 
to  the  first  two.    Hence  the  equations  of  the  line  become  — —  =  ' — —  '2=3. 

Geometrically  it  is  evident  that  the  two  points  lie  in  the  plane  2  =  3,  and  hence 
the  line  joining  them  also  lies  in  that  plane. 


THE  STKAIGHT  LINE  IN  SPACE  285 

9.  Find  the  equations  of  the  line  through  the  point  (—2,  4,  0)  which  is 

parallel  to  the  line  -  =  ^ = ,  and  reduce  them  to  the  answer. 

^  ^  -'^      Am.   x  =  -4z-2,y  =  -3z  +  -i. 

,«.^,  ,        ,     ,.        x  +  2      w  — 3      2-1       ,x-S      y      z  +  3 

10.  Show  that  the  lines  — ^—  = = and  =  -  =  — ^ 

,.     ,  6-3  2  2  6  3 

are  perpendicular. 

«• Q      ?/4-l      z 3 

11.  Find    the    angle    between    the    lines  = = and 

°  9  1  1 

X  +  2      V 7z  27r 

= =  -,  if  both  are  directed  upward.  Ans. 

12  1  3 

12.  Find  the  parametric  equations  of  the  line  passing  through  the  point 
(2,  —  3,  4)  whose  direction  cosines  are  proportional  to  1,  —  2,  and  2. 

Ans.   X  =  2  + 1/3,  y=-3-|p,  z  =  4  +  |p. 

13.  Construct  the  lines  whose  parametric  equations  are 

(a)x  =  2  +  |p,        2/  =  4-ip,     z  =  6  +  |p. 
(b)x=-3-f/),     ^  =  6-fp,     z  =  4  +  fp. 

14.  Find  the  distance,  measured  along  the  line  x  =  2  —  y\  p,  ?/  =  4  +  i|  p, 
z  =  —  3  +  y*3  p,  from  the  point  (2,  4,  —  3)  to  the  intersection  of  the  line  with 
the  plane  4x  —  i/ —  2z  =  6.  Ans.    1|. 

15.  Sliow  that  the  symmetric  equations  of  the  straight  line  become 

i  = i,  z  =  z,,  if  COS 7  =  0.    What  do  they  become  if  cos  a  =0  ? 

cos  a        cos  /3 

if  cos  /3  =  0  ? 

16.  Show  that  the  symmetric  equations  of  the  straight  line  become 
z  =  z^,  X  =  Xj,  if  cos  7  =  cos  o"  =  0.  What  do  they  become  if  cosa  = 
cos  /3  =  0  ?   if  cos  /3  =  cos  7  =  0  ? 

17.  Reduce  the  equations  of  the  following  lines  to  the  symmetric  form 
(IV). 

(a)  X  -  2  y  +  z  =  8,  2  X  -  3  y  =  13.  Ans. ^  =  -  = ^  • 

3  :;^  1 

Solution.  Find  the  equations  of  two  projecting  planes.  The  second 
plane  is  already  the  projecting  plane  on  XY.  Eliminating  x,  we  get 
y  —  2z  =—  S.    Now  in  the  two  projecting  planes  thus  found, 

(1)  2x-Sy-lS     and     y-2z=-3, 
solving  each  for  y  and  equating  results, 

,„,  2x-13_7/_22-3 

(2)  ^^______. 

Multiplying  the  numerators  through  by  I,  we  have  the  answer.. 


28G  NEAV  AXALYTIC   GEOMETRY 

Comparison  with  (IV)  gives  x^  =  Y,  j/j  =  0,  z^  =  |,  a  =  3,  6  =  2,  c  =  1. 
Hence  the  line  passes  through  (i/,  0,  |)  and  its  direction  cosines  are 
proportional  to  3,  2,  1. 

A  remark  here  is  important.  In  (IV),  Xj,  ?/j,  and  Zj  are  the  coordinates 
of  any  fixed  point  on  the  line.  Hence  for  a  given  line  the  numerators  in 
(IV)  may  be  quite  different.   For  example,  putting  z  =  0  in  (1),  we  find 

X  —  2      y  +  3      z 
X  =  2,  y  =—  S.    Hence  the  equations  = =  -  represent  the 

given  line  also.  Notice  that  in  equations  (IV)  the  coefficients  of  x,  y,  and 
z  must  be  unity.  This  explains  the  step,  after  deriving  (1),  of  removing  the 
2  from  the  2x  and  the  2z. 

(b)  4x-5?/  +  3z  =  3,  4x-5;/ +z  +  9  =  0.   Ans.   -  =  1^^1,2  =  6. 

5  4 

(c)  2x  +  z  +  5  =  0,  x  +  3z-5  =  0.  Ans.   z  =  3,  x  =  —  4. 

(d)  x  +  2y  +6z  =  5,  3x-2j/-10z  =  7.    Ans.   ?-^  = '/^  =  -  . 

(e)  3x-2/-2z  =  0,  6x-32/-4z  +  9  =  0.    Ans.   ^-^-=?,?/  =  9. 

(f)  3x  — 4j/ =  7,  X  +  3?/ =  11.  Ans.   x  =  5,  ?/ =  2. 

y    3  2 

(g)  2x  +  ?/  +  22  =  7,  X  +  31/  +  6z  =  11.       Ans. = ,  x  =  2. 

(h)  2x  — 3?/ +  z  =  4,  4x— 62/-Z  =  5.         Ans.    -^^-JL-,  z  =  l. 
(i)  3z  +  y  =  l,4z-Sy  =  10.  Ans.   y=-  2,  z  =  1. 


X  —  (X       y f) 

(])  X  =  rnz  +  a.,  y  =  nz  +  b.  Ans.    = 


1 


18.  Find  the  equations  of  the  line  passing  through  the  point  (2,  0,-2), 

J.    Q  y  Z    -1-    1 

which   is   perpendicular   to   each   of    the    lines =  -  = and 

'^:  =  ^±I  =  "-±!.  ,    %-2     i/%  +  2 

q  1  9  Ans.    =  -  = 

19.  Find  the  equations  of  the  line  passing  through  the  point  (3,  —  1,  2) 
which  is  perpendicular  to  each  of  the  lines  x  =  2z  —  1,  y  =  z  +  3,  and 
x_y_z  .    _    X  —  3_y  +  l_g  —  2 
2""3~4'  "^'        1      ~    -6  ~~i 

20.  Find  the  equations  of  the  line  through  P^^  (x^,  y^,  Zj)  parallel  to 


a  b  c  a  b  c 

(b)  X  —  7nz  +  a,  y  =  nz  +  b.  Ans. 


a;  —  x^  _  y  —  Vi  _  2 


m  71  1 


THE  STRAIGHT  LINE  IN  SPACE  287 

(0)  z  =  a,  y  =  mx  +  b.  Ans.   ^  = i,  z  =  z.. 

1  m 

(d)  A^x  +  B^y  +  C^z  +  D^  =  0,  A^x  +  B^y  +  C.,z  +  Z>2  =  0. 
Ans. 


x-x^       _        y-y^ 


B^C^  -  B^C,      C,A^  -  A^C^      A^B.,  -  A^B^ 

21.  Find  the  equations  of  the  line  passing  through  P^  (a;^,  j/^,  z^)  which 
is  perpendicular  to  each  of  the  lines 

X-X2  ^  y-y<2.  ^  Z-Z.2  ^^^  x-Xg  ^  y-y^,  ^  z-Zg  _ 
02  62  ^2     '  «3  ^3  Cg 

^oCg   —    63C2  fottg   —    CgOg  Og^g   —    ttg^g 

110.  Relative  positions  of  a  line  and  plane.  If  the  equations 
of  the  line  have  the  form  (IV),  and  if  Ave  substitute  the  values 
of  two  of  the  variables  given  by  (IV)  in  the  equation  of  the 
plane,  then  if  the  result  is  true  for  all  values  of  the  third  vari- 
able, the  line  lies  in  the  plane. 

We  next  easily  prove  the 

Theorem.  A  line  tvlwse  direction  angles  are  a,  /8,  and  y  and 
the  jilane  Ax  -\-  By  -\-  Cz  -\-  D  =  0  are 

(a)  parallel  when  and  only  when 

A  cos  a  -f  5  cos  y?  +  C  cos  y  =  0 ; 

(b)  perpendicular  when  and  only  when 

A.     _     B     _      C 

cos  a       cos  yff       cos  y 

Proof.  The  direction  cosines  of  a  perpendicular  L^  to  the 
plane  are  proportional  to  A,  B,  and  C. 

The  line  and  plane  are  parallel  when  and  only  when  the  line 
is  perpendicular  to  the  line  7.., ;  that  is,  when  and  only  when 
A  cos  a  ■\-  B  cos  ^-\-  C  cos  y  =  0. 

The  line  and  plane  are  perpendicular  when  and  only  when 
the  line  is  parallel  to  L^ ;  that  is,  when  and  only  when 
cos  a      cos  )8      cos  y 


288  NEW  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Show  that  the  line  = =  -  is   parallel   to  the  plane 

4x  +  2  2/  +  2z  =  9.  ^  ~^        ^ 

2.  Show  that  the  line  -  =  -  =  -  is  perpendicular  to  the  plane  3  x  + 
22/  +  7z  =  8.  3      2       7 

3.  Show  that  the  line  x  =  z  —  4,  y  =  2z  —  Z  lies  in  the  plane  2x  — 

4.  Find  the  equations  of  the  line  passing  through  (1,  —  6,  2  )  and  per- 
pendicular to  the  plane  2x  —  w  +  6z  =  0.         .         x  —  \      2/  +  6      z  —  2 

Ans.    = = . 

2-16 

5.  Show  that  the  lines  x  =  2z  +  l,  2/  =  32  +  2,  and  2x  =  z  +  2, 
3  2/  =  6  —  z  intersect,  and  find  the  equation  of  the  plane  determined  by 
them.  Ans.   20x  —  9y  —  13z  —  2  =  0. 

J. 2      w4-2      z 3 

6.  Show  that  the  line  = = lies  in  the  plane  2  x  + 

2i/-z  +  3  =  0.  ^  ~^  "* 

7.  Find  the  equations  of  the  line  passing  through  the  point  (3,  2,  —  6) 
which  is  perpendicular  to  the  plane  4x  —  2/  +  3z  =  5. 

x-3      y -2      z+6 


Ans. 


-1  3 


8.  Find  the  equations  of  the  line  passing  through  the  point  (4,  —  6,  2) 

which  is  perpendicular  to  the  plane  x4-2y  —  32  =  8. 

x-4      2/  +  6      z-2 

Ans.    = = 

12-3 

9.  Find  the  equations  of  the  line  passing  through  the  point  (—  2,  3,  2) 

which  is  parallel  to  each  of  the  planes  3x  —  ?/  +  z  =  0  and  x  —  z  =  0. 

.         x  +  2      y-d,      z-2 

Ans.    = = 

1  4  1 

10.  Find  the  equation  of  the  plane  passing  through  the  point  (1,  3,  —  2) 

which  is  perpendicular  to  the  line = = 

^  o  * —  1 

Ans.    2 X  +  5 1/  —  z  =  19. 

11,  Find  the  equation  of  the  plane  passing  through  the  point  (2,  —  2,  0) 
which  is  perpendicular  to  the  line  z  =  3,  y  =  2x  —  4.  Ans.   x  +  2y  ■\-  2  —  0. 


THE  STRAIGHT  LINE  IN  SPACE  289 

12.  Find  the  equation  of  the  plane  passing  through  the  line  x  +  2  z  =  4, 

J* Q       7/  _i_  4      z 7 

y  —  z  =  8  which  is  parallel  to  the  line = = 

2  3  4 

Ans.   a;  +  lOy- 8z- 84  =  0. 

13.  Find  the  equation  of  the  plane  passing  through  the  point  (3,  6,  —  12) 

which  is  parallel  to  each  of  the  lines  = = and 

z+2  3-132 

=  ——'2/ =  3.  Ans.    2x  +  Sy  —  z=zS6. 

14.  Find  the  equations  of  the  line  passing  through  the  point  (3,  1,  —  2) 
"which  is  perpendicular  to  the  plane  2x  —  y  —  5z  =  6. 

Ans.    2  =  —  I  z  -f  y-,  ?/  =  ^  z  +  |. 

,,    „,  ,         ,      ,.        X  — 2      ^  +  1        z  ,x  — 2      y  +  l      z 

16.  Show  that  the  lines  = = and  = =  - 

3  4-2  -18  2 

intersect,  and  find  the  equation  of  the  plane  determined  by  them. 

Ans.    14x  — 4y  +  13z  =  32. 

/g 2      7/4-3 

16.  Find  the  equation  of  the  plane  determined  by  the  line =  ~ 

z  —  1  " 

= and  the  point  (0,  3,  —  4).  Ans.   x  +  2y  +  2z  +  2  =  0. 

17.  Find  the  equation  of  the  plane  determined  by  the  parallel  lines 

x+1      V —2      z       ,x— 3      w+4     z— 1 

t^^  =  ^ =  Iand ^!LLZ  =  ^ — i.    Ans.   8x  + y- 26z  + 6  =  0. 

3  2  13  2  1  -T-y  -r 

18.  Find  the  equations  of  a  line  lying  in  the  plane x  +  Sy  —  2z  +  4^  =  0 

^ 4      ?/  +  2      z 2 

and  perpendicular  to  the  line = = at  the  point  where  it 

meets  the  plane. 

19.  Find  the  equations  of  a  line  tangent  to  the  sphere  x^  +  y"^  +  z^  =  9 
at  the  point  (2,  —  1,  —  2),  and  parallel  to  the  plane  x  +  3j/—  5z  —  1  =  0. 

20.  Find  the  equations  of  a  line  tangent  to  the  sphere  x^  +  y'^  +  z"^  =  9 
at  the  point  (2,  2,  —  1),  and  perpendicular  to  the  line = =  - . 

o  1  O 

21.  Find  the  equations  of  the  line  passing  through  Pi(Xj,  y^,  Zj)  which 
is  perpendicular  to  the  plane  Ax  +  By  +  Cz  +  D  =  0. 

A  B  C 

22.  Find    the    equation    of    the    plane    passing    through    tlic    point 


X  —  x,  _  ?/  —  2/2  _  z  —  z^ 
a  b  c 

Ans.    a(x  —  Xj)  +  6(y  —  ?/i)  +  c(z  —  Zj)  =  0. 


Pi(Xj,  2/j,  z,)  which  is  perpendicular  to  the  line 

a  b  c 


290  NEW  ANALYTIC  GEOMETRY 

23.  Find  the  angle  6  between  the  line ^-  = ^  = and  the 

plane  ^x+Bi,+  C.+  Z-  =  0.  «       ^„  %,  ^  ^^ 

^ns.    sin  0  =  

V^=2  4-  ir-i  +  C2  Va2  +  62  +  c2 

i/i/zi.  The  angle  between  a  line  and  a  plane  is  the  acute  angle  between  the 
line  and  its  projection  on  the  plane.  This  angle  equals  —  increased  or  decreased 
by  the  angle  between  the  line  and  the  normal  to  the  plane. 

24.  Find  the  equation  of  the  plane  pa.ssing  through  F^  (x.,,  2/3,  Zg)  which 

is  parallel  to  each  of  the  lines  ?Z1^  =  yjzll  =  ^  -  ~\  ^^^^^  -''■-'  2  ^  V-Vi 

a,  61  Ci  a„  62 


^2 

Ans.    {b^c^  -  Vi)  {x  -  x^)  +  {c^a^  -  a^c^)  {y  -  2/3)  +  {ajb^  -  aj>{)  (2  -  Zg)  =  0. 

25.  Find  the  condition  that  the  plane  A^x  +  B^y  +  C^z  +  D^  =  0  should 
be  parallel  to  the  line  A„x  +  B^y  +  G^z  +  D„  =  0,  A^x  +  B^y  +  C^z  +  D3  =  0. 

Ans.   A^{B^C^  -  bIc^)  +  B,  (C^g  -  ~C,A,)  +  C,  {A„B^  -  A,B„)  =  0. 

26.  Find  the  equation  of  the  plane  determined  by  the  point  P,  (x^,  ?/^,  z,) 
and  the  line  J-^x  +  B-^y  +  C^z  +  D^  =  0,  A^x  +  B^y  +  CgZ  +  I).,  =  0. 

^jts.    (^2^1  +  So2/i  +  <^22i  +  -D2)  (^ia;  +  Bi2/  +  C^z  +  X>,) 

=  {A^x^  4-  -Ki2/i  +  C^Zj  +  Di)  (^„x  +  i^22/  +  C^z  +  X»2). 

27.  Find  the  equation  of  the  plane  determined  by  the  intersecting  lines 

«!  6j  Cj  tto  ft.,  c., 

^ns.    (ft^Co  -  62C1)  (X  -  Xj)  +  (Ci«2  -  C2ai)  (2/ -  ?/i)  +  {a A  "  «2'^i)  (- -  ^i)  =  0. 

28.  Find  the  equation  of  the  plane  determined  by  the  parallel  lines 

x-^i  ^  y-Vx  ^  ^-^1  and  ^~^'^  =  '^  ~  ''-  =  ^~~2  . 
a  b  c  a  b  c 

Ans.    [(2/1  -  2/2)  c  -  (z,  -  Z2)  b]x+  [(Zi  -  Za)  «  -  (^1 "  ^2)  c]  ^ 
+  [(-^1  -^2)^-  iVi  -  Vi)  «]  2  +  (2/1^2  -  2/2^1)  « 
+  (z,X2  -  ZoX^)  b  +  (X12/2  -  a^oJ/i)  c  =  0. 

29.  Find  the  conditions  that  the  line  x  =  mz  -\-  a,y  =  nz  +  b  should  lie 
in  the  plane  Ax  +  By  +  Cz  +  D  =  0. 

Ans.    Aa  +  Bb  +  D  =  0,  Am  +  Bn  +  C  =  0. 

30.  Find  the  equation  of  the  plane  passing  through  the  line 

y-y-i     z  —  z,    ^.  ,  .         „  w   .,    T     a;  — X,     y  —  y^     z  —  z^ 

—  - — -^  = 1  which  is  parallel  to  the  line =  — ; — -  = -■ 

bi  Ci  «2  ^2  ^2 

Ans.  (61C2  -  62C1)  {x-x^)  +  (Cia2  -  f otti)  {y -  2/0  +  (ai&2  -  «2^)  (^  "  ^i)  =  0- 


CHAPTER  XVII 

SPECIAL  SURFACES 

^  111.  In  this  chapter  we  shall  consider  spheres,  cylinders, 
and  cones*  (surfaces  considered  in  elementary  geometry),  and 
surfaces  which  may  be  generated  by  revolving  a  curve  about 
one  of  the  coordinate  axes,  or  by  moving  a  straight  line. 

112.  The  sphere.    We  begin  Avith  the 

Theorem.  Tlie  equation  of  the  sphere  tvhose  center  is  the  point 
(a,  (3,  y)  and  whose  radius  is  r  is 

(I)  {X  -  ay  +  (y  -  py  +  (z  -  yy  =  r\ 

Proof.  Ijct  P  (x,  y,  z)  be  any  point  on  the  sphere,  and  denote 
the  center  of  the  sphere  by  C.  Then,  by  dehnition,  PC  =  r. 
Sul)stituting  the  value  of  PC  given  by  the  length  formula, 
and  squaring,  we  obtain  (I).  q.  e.  d. 

When  (I)  is  multiplied  out,  it  is 

that  is,  it  is  in  the  form 

^■'^  -\- y'^ -V  z^ -\-  Gx -\-  liy  +  Iz  +  A'  =  0. 

The  question  now  is.  When  is  the  locus  of  tills  equation  a 
sphere  ? 

To  answer  this,  collect  the  terms  thus  : 

(a:^  +  Gx)  +  (//2  4-  ///y)  +  («'  +  /-)  =  -  K. 

*In  analytic  geometry  the  terms  "sphere,"  "cylinder,"  and  "cone  "  are 
usually  used  to  denote  the  spherical  surface,  cylindrical  surface,  and  conical 
surface  of  elementary  geometry,  and  not  the  solids  bounded  wholly  or  in  part 
by  such  surfaces. 

291 


292  NEW  ANALYTIC  GEOMETRY 

Completing  the  squares  within  the  parentheses,  we  obtain 
(x  +  1  Oy  +(!/  +  h_  Hf  +{z  +  \  If  =  \  {G-'  +  7/2  _^  /2  _  4  A'). 

Comparing  with  (I),  we  have  at  once  the 

Theorem.    The  loms  of  an  equation  of  tJie  foiin 
(II)  x^  +  z/2  4-  z2  +  Gat  +  fii/  4-  Iz  +K  =  0 

Is  determined  as  foUoivs  : 

(a)    IVhen  6''^+  7/^-1-  Z^—  4  K  >  0,  the  locus  is  a  sj^here^whose 
center  is  I  —  -^j  ~  "o  '  ~"  o  )  ^^^  whose  radius  is 


r  =  ^  VG'-  +  7/-  +  7'-4A:. 

(b)  When  6'^+  //'+  J^—  4  7v  =  0,  the  locus  is  the  point-sphere* 

~Y     ~    2'     ^2y 

(c)  When  G'-  +  77"  +  7^  —  4  7v  <  0,  there  is  no  locus. 

In  numerical  examples  it  is  recommended  that  the  theorem 
be  iiot  used,  but  that  the  squares  be  completed  as  in  the  proof, 
and  the  center  and  radius  be  found  by  comparison  with  (I). 


EXAMPLE 
What  is  the  locus  of  the  equation 

x2+2/2  4.22_2x  +  32/  +  l  =  0? 

Solution.    Collecting  terms, 

(x2-2x)  +  (2/'  +  32/)  +  z2  =  -l. 
Completing  the  squares, 

(x2-2x  +  l)  +  (2/2+32/  +  |)  +  2^  =  -l  +  1+1, 

or  (X-l)2+(y+|)-2+z2  =  | 

This  equation  is  in  the  form  (I);  r=|,  cr  =  1,  /3  =—  |,  7  =  0.    That  is, 
tlie  locus  is  a  sphere  of  radius  |  and  center  (1,  —  |,  0). 

*  That  is,  a  poiut  or  sphere  of  radius  zero. 


SPECIAL  SURFACES  293 

PROBLEMS 

1.  Find  the  equation  of  the  sphere  whose  center  is  the  point 

(a)  (a,  0,  0)  and  whose  radius  is  a.        Ans.   x^  +  y^  +  z^  —  2  ax  =  0. 

(b)  (0,  /3,  0)  and  whose  radius  is  (3.  Ans.   x^  +  y^  +  z'^—  2/3?/  =  0. 

(c)  (0,  0,  7)  and  whose  radius  is  7.  Ans.    x^  +  y^  +  z^  —  2  yz  =  0. 

2.  Determine  the  nature  of  the  loci  of  the  following  equations  and  find 
the  center  and  radius  if  the  locus  is  a  sphere,  or  the  coordinates  of  the 
point-sphere  if  the  locus  is  a  point-sphere. 

(ay'j24.2/2+z2_6x  +  4z  =  0.  (c)  x^+y^+ z^+ 4x  —  z+  7  =  0. 

(b)  'x^  +  y^  +  z^+  2x  -  4y  -  5  =  0.    (d)  x^+  y^+  z^-12x  +  6y  +  4z=0. 

3.  Where  will  the  center  of  (II)  lie  if 

(a)G  =  0?  (c)  7  =  0?  (e)  11=1  =  0? 

(b)fl"  =  0?  {d)G  =  n  =  Of  {i)I=G  =  0? 

4.  Prove  that  each  of  the  following  loci  is  a  sphere,  and  find  its  radius 
and  the  coordinates  of  its  center. 

(a)  The  distance  of  a  point  from  the  origin  is  proportional  to  the  square 
root  of  the  sum  of  its  distances  from  the  three  coordinate  planes. 

(b)  The  sum  of  the  squares  of  the  distances  of  a  point  from  two  fixed 
points  (2,  4,  —  8)  and  (-  4,  0,  2)  is  equal  to  104.  _ 

Ans.    ct  =  —  1,  /3  =  2,  7  =  -  3,  r  =  Vl4. 

(c)  The  distance  of  a  point  from  the  origin  is  half  its  distance  from 
the  point  (3,  —  6,  9). 

(d)  The  distance  of  a  point  from  the  point  (7,  1,  —  3)  is  twice  its  dis- 
tance from  the  point  (—  |,  —  2,  |).  /— — - 

Ans.    a=—  4,  /3=—  3,  7  =  1,  r  = • 

(e)  The  sum  of  the  squares  of  the  distances  of  a  point  from  the  three 
planes  -x  +  2y  +  2z  —  l  =  0,2x  —  y+2z  —  l  =  0,2x  +  2y—z  —  l=0 
is  unity. 

5.  Show  that  a  sphere  is  determined  by  four  conditions  and  formulate 
a  rule  by  which  to  find  its  equation. 

6.  Find  the  equation  of  a  sphere  passing  through  the  three  points  in 
any  one  of  the  following  columns  and  through  a  fourth  point  selected 
from  the  other  two. 

^(-1,-1,1).  -D (0,0,1),  G(0,  -4,  5), 

B{-1,-  3,  1),  E(3,  0,  2),  H{2,  -  4,  5), 

C(-l,  -4,  4);  i^(2,  0,1);  7(3,-1,5). 

Ans.    x^  +  y^  +  z^  —  2x  +  4y  —  6z  +  b  =  0. 


294  NEW  ANALYTIC  GEOMETRY 

7.  Find  the  equation  of  a  sphere  which 

(a)  has  the  center  (3,  0,  —  2)  and  passes  through  (1,  6,  —  6). 

Ans.   x^+  y'^  +  z^—  Qx  +  4 z  —  36  =  0. 

(b)  passes  through  the  points  (0,  0,  0),  (0, 2,  0),  (4,  0,  0),  and  (0,  0,  —  6). 

Ans.   x^+ y^  + z^—4x  —  2y  +  6z  =  0. 
( c  )  is  concentric  with  the  sphere  x^  +  y^  +  z^  —  6x  +  4-;;  =  0  and  passes 
througli  the  point  (3,  1,  0). 

(d)  has  the  line  joining  (4,  —  6,  5)  and  (2,  0,  2)  as  a  diameter. 

( e )  lias  the  center  (2,  2,  —  2)  and  is  tangent  to  the  plane  2x  +  y  — 
3z  +  2  =  0. 

(  f  )  has  a  unit  radius  and  is  tangent  to  each  of  the  coordinate  planes 
in  the  first  octant. 

(  g )  passes  through  the  three  points  (1, 0,2),  (1,3, 1),  and  (—3,0,0)  and 
has  the  center  in  the  A'Z-plane.      Ans.   x^  +  y"^  +  z^  —  2x  +  6z —  15  =  0. 

(h)  passes  through  the  three  points  (1,  —  3, 4),  (1,  —  5,  2),  and  (1,  —  3, 0) 
and  has  its  center  in  the  plane  x  +  ?/  +  z  =  0. 

Ans.  x-  +  y^  +  z"  —  2 x  +  6y  —  4 z  +  10  —  0. 

( i )  has  its  center  on  the  I'-axis  and  passes  through  the  points  (0,  2,  2) 
and  (4,  0,  0).  Ans.  x"  +  y"  +  z^  +  iy -16  ^0. 

( j )  passes  through  the  points  (1,  5,  —  3)  and  (—  3,  0,  0),  and  whose 
center  lies  on  the  line  of  intersection  of  the  planes  3x  +  y  +  z  =  0,  x  + 
2y  —  l=0.  Ans.    x"^  +  y~  +  z- -  2x  +  6z —  15  =  0. 

(k)  is  tangent  to  the  three  coordinate  planes  and  to  the  plane 
6a;  +  22/  +  3z  —  4  =  0.         Ans.   x^  +  y-  +  z- -  2x  -  2y -2z  —  2  =  0. 

( 1 )  has  its  center  at  (3,  1, 1)  and  is  tangent  to  the  sphere  x^  +  y'^  + 
22_2x-4y  +  2z  +  2  =  0.     Ans.   x^  +  y^  +  z- -6x  —  2y  -  2z +10  =  0; 

x^  +  y^  +  z-  —  6x  —  2y  —  2z  —  U  =  0. 

(m)  passes  through  the  points  (1, 1,  0),  (0,  1,  1),  and  (1,  0,  1)  and  whose 
radius  is  11.  -4ns.   x^  +  2/'^  +  z- —  14x  —  14?/ —  14z  +  26  =  0. 

( n)  is  tangent  to  the  plane  x+y  —  z  +  l  =  0?it  the  point  (3,  —  2,  2) 
and  has  its  center  in  the  XF-plane. 

(o)  passes  through  the  three  points  (2,  0, 1),  (2,  —  1,  0),  and  (1,  —  1,  1) 
and  is  tangent  to  the  plane  2x  +  2y  —  z  +  2  =  0. 

Ans.   x"^  +  y'^  +  z^  —  4  x  +  2  y  —  2  z  +  5  =  0. 

( p )  passes  through  the  intersection  of  the  two  spheres  x^  +  y^  +  z^  — 
6x  =  0,  x^  +  y^  +  z^  +  9y  —  5z  —  7  =  0,  and  through  the  point (0,  1,  1). 

8.  Find  the  equations  of  the  tangent  plane  and  the  normal  line  to 
the  sphere  x^  +  y^  +  z^ —  14  =  0  a.t  the  point  (3,  -  2,  1). 

9.  Find  the  equations  of  the  tangent  plane  and  normal  line  to  the 
sphere  x^  +  j/2  +  j;2  _  2x  +  4?/  —  6z  +  5  =  0  at  the  point  (3,  —  4,  2). 


SPECIAL  SURFACES 


295 


10.  Find  the  equations  of  tlie  planes  tangent  to  tlie  sphere  x^  +  y^  -\-z'^  — 
lOx  +  5y—2z  —  24  =  0  at  the  points  where  it  intersects  the  coordinate  axes. 

11.  Find  tlie  equation  of  a  sphere  inscribed  in  the  tetrahedron  formed 
by  any  four  of  the  following  planes  : 

14j;+    52/-2Z-168  =  0,  lOx  +  Uy  +  2z+    88  ==  0, 

14x—    5y  +  2z+28  =  0,  2x-      y-2z+    12  =  0, 

10a;-ll2/4- 2z+    33  =  0,  2x-      y  +  2z+      8  =  0. 

12.  Find  the  equation  of  the  smallest  sphere  tangent  to  the  two  spheres 
x^  +  y^  +  z^ -2x-6y  +  1  =  0,  x"^  +  y-  +  z^  +  6x  +  2y-  4:Z+  5  =  0. 

Ans.   x^  +  i/^  +  z-  +  2  x  —  2  y  —  2  z  +  S  =  0. 

113.  Cylinders.  A  surface  which  is  generated  by  a  straight  line 
which  moves  parallel  to  itself  and  intersects  a  given  fixed  (iurve 
is  called  a  crjlinde):  The  fixed  curve  is  called  the  directrix.  We 
now  consider  equations  whose  loci  are  cylinders. 


EXAMPLES 

1.  Determine  the  nature  of  the  locus  of  y"^  =  4nX. 

Solution.    The  intersection  of  the  surface  with  a  plane  x  =  k,  parallel 
to  the  FZ-plane,  is  the  pair  of  lines 
(1)  x  =  k,    y  =  ±2\^, 

which  are  parallel  to  the  Z-axis.  If  A:  >  0,  the  locus  of  equations  (1)  is  a 
pair  of  lines  ;  if  A;  =  0,  it  is  a  single  line  (the  Z-axis)  ;  and  if  fc  <  0, 
equations  (1)  have  no  locus. 

Similarly,  the  intersection 
with  a  plane  y  =  k,  parallel 
to  the  Z  J-plane,  is  a  straight 
line  whose  equations  are 

x  =  \k'',     y  =  k, 

and  which  is  therefore  par- 
allel to  the  Z-axis. 

The  intersection  with  a 
plane  z  =  k  parallel  to  the 
A'F-plaue  is  the  parabola 

z  =  k,     2/2  =  4  X. 

For  different  values  of  k  these  parabolas  are  equal  and  placed  one 
above  another.  The  surface  is  therefore  a  cylinder  whose  elements  are 
parallel  to  the  Z-axis  and  intersect  the  parabola  y^  =  ix,    z  =  0. 


296 


NEW  ANALYTIC  GEOMETRY 


It  is  evident  from  Ex.  1  that  the  locus  of  any  equation  which 
contains  but  two  of  the  variables  x,  y,  and  z  will  intersect 
planes  parallel  to  two  of  the  coordinate  i)lanes  in  one  or  more 
straight  lines  parallel  to  one  of  the  axes,  and  planes  parallel 
to  the  third  coordinate  plane  in  congruent  curves.  Such  a  sur- 
face is  evidently  a  cylinder.    Hence  the 

Theorem.  The  locus  of  an  equation  in  which  one  \iariahle  is 
lacking  is  a  cylinder  wliose  elements  are  j)arallel  to  the  axis 
along  which  that  variable  is  tneasured. 

The  student  should  not  infer  from  this  statement  that  the 
equations  of  all  cylinders  have  one  variable  lacking.  In  case 
the  elements  are  inclined,  all  three  variables  will  appear  in  the 
equation.    This  is  illustrated  by  the  following  example : 

2.  Determine  the  nature  of  the 
locus  of 

X-  +  2  .cz  +  z^  =  1  —  y"^. 

The  intersection  of  this  locus 
by  the  i^lane  y  =  k  is 

y  =  k,     X  +  z  =  zb  Vl  — ^-, 

a  pair  of  parallel  lines  whose 
direction  is  independent  of  k.  In 
fact,  the  direction  cosines  of  these 
lines  are  proportional  to  —  1, 0, 1 ; 
that  is,  they  are  parallel  to  the 
line  joining  the  point  (—1,  0,  1) 
to  the  origin.  "We  conclude  then 
that  the  surface  is  a  cylinder.  To 
construct  the  surface,  draw  its  traces  and  pass  lines  through  them  hav- 
ing the  above  direction.   The  trace  in  the  FZ-plane  is  the  circle 

y"  +  z^  =  l; 
in  the  XY-plane,  the  circle 

x2  +  2/2  =  1. 

It  is  evident  from  Ex.  2  that  in  order  to  prove  that  a  surface  is 
cylindrical  it  is  only  necessary  to  find  a  system  of  planes  which  cut 
from  it  a  system  of  parallel  lines. 


SPECIAL  SURFACES  297 

PROBLEMS 

1.  Determine  the  nature  of  the  following  loci,  and  discuss  and  con- 
struct them  : 

(a)  x2  +  t/2  =  36.  (f )  z^  +  x^  =  f\ 

(b)  x2  +  ?/  =  3x.  (g)  x^  +  6y  =  0. 

(c)  x^-z^  =  16.  ,         (h)  yz-4  =  0. 

(d)  2/2  +  4 z2  =  0.  (i)  2/2  +  2-4  =  0. 

(e)  x2+22/-4  =  0.  (j)  2/2_a.3^0. 

2.  Find  the  equations  of  the  cylinders  whose  directrices  are  the  folJow- 
ing  curves  and  whose  elements  are  parallel  to  one  of  the  axes : 

(a)  2/2  +  22  _  4  ?/  =0,  X  =  0.  (c)  b-^x^  -  a^y"-  =  a?b^,  2  =  0. 

(b)  22  +  2  X  =  8,  2/  =  0.  (d)  ?/2  +  2p2  =  0,  X  =  0. 

3.  Prove  that  the  following  loci  are  cylinders.  Discuss  and  construct 
them,     (a)  X  +  ?/  -  22  =  0.  (d)  x2  -  4  (2  +  ^)  +  8  =  0. 

(b)  X2  +  ?/2  —  1  =  0.  (e)  x2  +  2x2/  +  Ip-  =  2- 

(c)  y2  =^  3x  +  2.  (f )  x2  -  2x2/  +  ?/2  =  1  -  z^- 

4.  A  point  moves  so  that  its  distance  from  a  fixed  point  is  always  equal 
to  its  distance  from  a  fixed  line.  Prove  that  the  locus  is  a  parabolic 
cylinder. 

6.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  two  intersecting  perpendicular  lines  is  constant.  Prove  that  the 
locus  is  a  hyperbolic  cylinder. 

6.  A  point  moves  so  that  the  sum  of  its  distances  from  two  planes  is 
equal  to  the  square  of  its  distance  from  a  third  plane.  The  three  planes 
are  mutually  perpendicular.    Prove  that  the  locus  is  a  parabolic  cylinder. 

7.  A  point  moves  so  that  the  sum  of  its  distances  from  two  planes  is 
equal  to  the  square  root  of  its  distance  from  a  third  plane.  Prove  that 
the  locus  is  a  parabolic  cylinder  when  the  three  planes  are  mutually 
perpendicular. 

114.  The  projecting  cylinders  of  a  curve.  The  cylinders  whose 
elements  intei'sect  a  given  curve  and  are  parallel  to  one  of  the 
coordinate  axes  are  called  the  projecting  cylinders  of  the  curve. 
The  equations  may  be  found  by  eliminating  in  turn  each  of  the 
variables  a-,  y,  and  z  from  the  equations  of  the  curve,  l^'or  if  we 
eliminate  z,  for  example,  the  result,  by  the  preceding  section,  is 


298 


NEW  ANALYTIC  GEOMETRY 


the  equation  of  a  cylinder  which  passes  through  the  curve,  since 
values  of  a-,  y,  and  ;;;  which  satisfy  each  of  two  equations  satisfy 
an  equation  obtained  from  them  by  eliminating  one  variable. 

The  equations  of  two  of  the  projecting  cylinders  may  be 
conveniently  used  as  the  equations  of  the  curve.*  Hence  the 
problem  of  constructing  the  original  curve  reduces  to  that  of 
constructing  the  curve  of  intersection  of  two  cylinders  whose 
elements  are  parallel  to  the  coordinate  axes.  The  method  is 
illustrated  in  the  following  examples. 


EXAMPLES 

1.  Construct  the  curve  of  intersection  of  the  two  cylinders 
x"-  +  1/2  -  2  ?/  =  0,     ?/  +  z2  _  4  =  0. 

Solution.    Draw  the  trace  of  each  cylinder  on  the  coordinate  plane  to 
v^hich  its  elements  are  pei'pendicular.   Then  consider  a  plane  perpendicu- 


F 

\ 

() 

/ 

A 

i 

h 

'/ 

^'    ' 

A 

<- 

c 

1 
1 

E 

^ 

G 

lar  to  the  coordinate  axis  to  which  the  elements  of  neither  cylinder  are 
parallel.    In  this  case  such  a  plane  is  ?/  =  fc.    Let  this  plane  intersect  the 

*  In  general,  the  equations  of  a  curve  may  be  replaced  by  any  two  inde- 
pendent equations  to  which  they  are  equivalent ;  that  is,  by  two  independent 
equations  which  are  derived  by  combining  the  given  equations. 


SPECIAL  SURFACES 


29^ 


axis  at  the  point  K.  It  will  intersect  the  traces  at  the  points  J^,  B,  C,  and  I). 
Through  each  of  these  points  will  pass  an  element  of  the  corresponding 
cylinder,  all  four  elements  lying  in  this  plane.  The  points  of  intersection 
E,  F,  G,  and  H  of  these  elements  are  points  on  the  curve  of  intersection 
of  the  two  cylinders.  By  taking  several  positions  of  the  plane  y  =  k,  we 
obtain  a  sufficient  number  of  points  to  construct  the  entire  curve  as  shown 
in  the  second  figure  on  page  298. 

2.  Construct  the  curve  whose  equations  are 

2y^  +  z^  +  4x  =  4z,     y"^  +  Sz^  -  8x  =  12z. 

Eliminating  x,  y,  and  z  in  turn,  we  obtain  the  equations  of  the  project- 
ing cylinders 

2/2  +  z2  _  4 2^     z'^  —  4 X  =  4 r,     y'^  -\-  Ax  —  0. 


The  figure  shows  the  first  and  third  of  these  cylinders,  intersecting 
in  the  original  curve  constructed  by  the  method  explained  in  the 
previous  example. 

It  is  usually  wise  to  deduce  the  equations  of  all  three  of  the  projecting 
cylinders,  for  it  may  be  that  two  of  them  are  distinguished  for  simplicity 
and  hence  are  most  convenient  to  construct. 

If  the  curve  lies  in  a  plane  parallel  to  one  of  the  coordinate 
planes,  then  two  of  its  projecting  cylinders  coincide  with  the 
plane  of  the  curve,  or  part  of  it. 

For  a  straight  line  the  projecting  cylinders  are  the  j)roject- 
ing  planes. 


3P0  NEW  ANALYTIC   GEOMETRY 

PROBLEMS 

1.  Construct  the  curve  in  which  the  following,  in  each  case  a  plane 
and  a  cylinder,  intersect : 

2.  Construct  the  curve  in  which  the  following  pairs  of  cylinders  intersect : 

rx2-42/  =  0,  rx2+2/2  =  25, 

^'\y-^+4z  =  0.  ^  '  \5z  +  y^  +  10y  =  0. 

ri/2  +  4  z  =  0,  r?/-  +  z'  -  36  =  0, 

^    '    |x'2  +  2/2 -  4  =  0.  ^°^    \x2  +  ^2_7y^0. 

rx2-  9z  +  36  =  0,  r2/2  +  z2_  36  =  0, 

^^^  |x2  +  2/2  _  36  =  0.  ^  ^  |x2  +  2/2  -  5  y  =  0. 

r2/-^  +  4.  =  0,  r2/2+x2_36  =  0, 

^  ^  \x2+2/2-42/  =  0.  ^  ^  \z2+2/2-62/  =  0. 

rx2+z2_25=0,  rz2/  =  12, 

(^)  I2/2  -  z  =  0.  ^^^  |x2  +  2/2-  72/  +  6  =  0. 

3.  Find  the  equations  of  the  projecting  cylinders  of  the  following  curves 
and  construct  the  curve  as  the  intersection  of  two  of  these  cylinders  : 

(a)  x2  +  2/2  +  z2  =  25,  x2  +  4  y'^  -  z^  =  0. 

(b)  a;2  +  4  2/2 -  z2  =  16,  ix^  +  y^  +  z^  =  16. 

(c)  x2+  2/2=  4z,  x'-—  2/2  =  8 z. 

(d)  x2  + 22/2+ 4z2=  32,  x2  + 42/2  =  4z.  f  2/2- x2+ 2^2+ 72/- 72  =  0, 

(e)  2/2  +  z,c  =  0,  2/-  +  2x  +  2/  -  2  =  0.        ^  ^^   |x2  _  z2  _  7  ^^  ^  36  =  0. 

rx2_  102/ -5z- 25  =  0,  r2x2+2/2-9z  =  0, 

^  ^  V-+  2  2/2+  5z+102/-25  =  0.       ^^'  l?/^  +  9z-72  =  0. 

rx2+ 22/2  + 4z- 4  =  0,  f2x2  +  22/2 +  Z2/- 142/ =  0, 

^°^  |2x2+ 52/2+ 12z- 8  =  0.  ^-"^  |x2  +  2/2  +  2z2/-72/-18  =  0. 

4.  A  point  is  two  units  from  the  Z-axis  and  the  sum  of  its  distances 
from  the  A'F-plane  and  the  FZ-plane  is  equal  to  its  distance  from  the 
ZX-plane  increased  by  2.    Construct  its  locus. 

5.  A  point  is  equidistant  from  the  Z-axis  and  the  A'F-plane,  and  its 
distance  from  the  origin  is  equal  to  its  distance  from  the  FZ-plane 
increased  by  2.    Construct  the  locus. 

115.  Parametric  equations  of  curves  in  space.  If  the  coordi- 
nates X,  j/,  and  2;  of  a  point  P  in  space  are  functions  of  a  variable 
parameter,  then  the  locus  of  P  is  a  curve  (compare  Art.  80). 


SPECIAL  SURFACES 


301 


For  example,  if 

(1)  x  =  ii^     2/  =  l-2i,     2  =  3^3+2, 

where  tisa  variable  parameter,  then  the  locus  of  {x,  y,  z)  is  a  curve  in  space. 
This  curve  may  be  drawn  by  assuming  values  for  t,  computing  x,  y,  and  z, 
plotting  the  points,  and  then  joining  these  points  in  order  by  a  continuous 
curve.    Equations  (1)  are  called  the  parametric  equations  of  the  curve. 

The  equations  of  the  projecting  cylinders  of  the  curve,  the  locus  of 
(1),  result  when  the  parameter  t  is  eliminated  from  each  pair  of  the 
equations.    Thus,  taking  the  first  two, 

(2)  x  =  lt^     y  =  l-2t, 

we  find  from  the  second,  t  =  l(^  —  y),  and  substituting  in  the  first, 

(3)  4a;  =  i(l-?/)2,     or  (?/ -  1)2- I6x  =  0, 
and  the  locus  lies  on  this  parabolic  cylinder. 

Similarly,  eliminating  t  from  the  first  and  third  equations  of  (1), 
x  =  \t-,     2  =  3(3+2, 
we  obtain  the  cubic  cylinder 

(4)  (2 -2)2  =.5  76x3. 

Hence  the  curve  (l)  is  the  curve  of  intersection  of  the  cylinders  (3) 
and  (4). 

In  some  cases  it  is  convenient  to  find  the  equations  of  a  curve  in  space 
by  using  a  parameter. 

EXAMPLE 

Equations  of  the  helix.  A  point  moves  on  a  right  cylinder  in  such  a  man- 
ner that  the  distance  it  moves  parallel  to  the  axis  varies  directly  as  the 
angle  it  turns  through  around  the  axis. 
Find  the  equations  of  the  locus. 

Solution.  Choose  the  axes  of  coordi- 
nates so  that  the  ecjuation  of  the  cylin- 
der is 

(5)  X-  +  y-  =  a", 

as  in  the  figure. 

Let  Pq  on  OX  be  one  position  of  the 
moving  point,  and  P  any  other  position. 
Then,  by  definition,  the  distance  NP 
(=  z)  varies  as  the  angle  XON  {=6); 
that  is,  z  =  bd,  where  6  is  a  constant.  Furthermore,  from  the  figure, 
X  =  OM  =  ON  cosd  =  a  cos 0, 
y  =  MN  =  ON  sin  0  =  o  sin  0. 


302  NEW  ANALYTIC   GEOMETRY 

Hence  the  equations  of  the  helix  are  : 

(6)  X  =  a  cos  0,     y  =  asind,     z  =  bd, 
where  ^  is  a  variable  parameter.    Ans. 

Eliminating  0  from  the  first  two  of  equations  (6),  we  obtain  (5),  as 
we  should. 

Given  the  equations  of  the  projecting  cylinders,  to  find  para- 
metric equations  for  the  curve.  It  was  shown  in  Art.  81  that 
an  indefinite  number  of  parametric  equations  could  be  obtained 
for  the  same  plane  curve.  The  same  statement  holds  for  space 
curves,  as  illustrated  in  the  following  example. 

EXAMPLE 

Find  parametric  equations  for  the  curve  of  intersection  of  the  sui'faces 
(see  Example  2,  Art.  114), 

2^/2 +.2- +  4x  =  42,     ?/- +  322- 8a;  =  12z. 
Solution.   The  projecting  cylinders  are 

(7)  2/2 +  2^  =  42,     z^  —  \x  =  \z,     2/2+4a;  =  0. 

If  we  assume  y  =  2t,  then  the  last  equation  will  give  x  =—  t^.  From 
either  of  the  other  two  cylinders  we  find 

z  =  2±2  Vl  -  i2. 
Hence  the  given  curve  is  the  locus  of 

(8)  X  =-f\     y  =  2t,     z  =  2±2  Vl  -  t-. 

Other  parametric  equations  result  when  we  set  one  of  the  coordinates 
in  (7)  equal  to  some  other  function  of  a  parameter.  The  aim  is,  of  course, 
to  find  simple  parametric  equations.  The  method  adopted  must  depend 
upon  the  given  problem. 

PROBLEM 

Find  simple  parametric  equations  for  the  curves  of  Problems  2  and  3, 
p.  300. 

Ans.  For  Problem  2.   (a)  x  =  2t,  y  =  t^,  2  =  —  |  <*. 

(b)  x  =  2cos^,    y  =  2smd,     z=—s\n^O. 

(c)  X  =  6  cos  0,    y  =  6  sin  0,     z  =  4(1  +  cos^  0). 

116.  Cones.  The  surface  generated  by  a  straight  line  turning 
around  one  of  its  points  and  intersecting  a  fixed  curve  is  called 
a.6'07ie. 


SPECIAL  SURFACES 


303 


EXAMPLE 

Determine  the  nature  of  the  locus  of  the  equation  16  x^  +  y^  —  z^  =  0. 

Solution.    Let  Pj  (x^,  y^,  z^)  be  a  point  on  a  curve  C  on  the  surface 
in  which  the  locus  intersects  a  plane,  for  example  z  =  k.    Then 
(1)        16x{  +  y^  -z^  =  0,     Zi  =  k. 

Now  the  origin  O  lies  on  the  surface.  "We 
shall  show  that  the  line  OPj  lies  entirely  on 
the  surface.    The  direction  cosines  of  OP.^ 

are  -^,  -^,  and  -^ ,  where  p?  =  x?  +  yr  +  z? 
Pi    Pi  Pi  A-i         1         1        1 

=  OP^.  Hence  the  coordinates  of  any  point 

on  OP^  are,  by  (II),  Art.  109, 


(2) 


X  =  ~p, 
Px 


Vi 


z  =  ^p. 
Pi 


Substituting  these  values  of  x,  ?/,  and 
2  in  the  left-hand  member  of  the  given 
equation,  we  obtain 

P'l    ' 


(3) 

^Q^iP'^y<p 

pI          Pi 

or  also 

0 

(4) 


-2(lQx{+y{-z{). 


But  from  the  first  of  equations  (1)  the 
expression  in  the  parenthesis  in  (4)  equals 
zero.  Hence  the  product  in  (4)  also  van- 
ishes for  any  value  of  p.  This  means  that 
every  point  (x,  y,  z)  on  the  line  (2)  lies  on  the  surface,  that  is,  the  entire  line 
lies  on  the  surface.   Hence  the  surface  is  a  cone  whose  vertex  is  the  origin. 

The  essential  thing  in  the  above  solution  is  that  (4)  may  be 
obtained  from  the  first  of  equations  (1)  by  multiplying  liy  a 

P 
power  of  —  •    This  may  be  done  whenever  the  equation  of  the 

surface  is  homogeneous  *  in  the  variables  x,  y,  and  z.    Hence  the 
Theorem.    The  locus  of  an  equation  which  is  homogeneous  in 
the  variables  x,  ]/,  and  z  is  a  cone  whose  vertex  is  the  origin. 

*  An  equation  is  homogeneous  iu  x,  y,  and  z  when  all  the  terms  in  the 
ei]  nation  are  of  the  same  degree. 


304  NEW  ANALYTIC  GEOMETRY 

To  construct  the  locus  of  the  equation  of  a  cone,  find  the 
intersection  of  the  cone  with  a  suitably  chosen  plane  parallel 
to  one  of  the  coordinate  planes,  construct  this  plane  curve,  and 
then  draw  the  elements  from  the  points  on  this  curve  to  the 
vertex  of  the  cone. 

Thus  in  the  figure  for  the  preceding  example,  the  cone  is 
cut  by  the  plane  s  =  8,  and  the  curve  of  intersection,  namely 
the  ellipse  16  cc^  +  y^  —  64  =  0,  is  drawn  in  this  plane. 

PROBLEMS 

1.  Determine  the  nature  of  the  following  loci,  and  discuss  and  con- 
struct them  : 

(a)  x^-ii'-\-  36  z-  =  0.  (e)  x'^  +  9  ;/-  -  4  z2  -  o. 

(b)  ?/2-16x2+ 4z2  =  0.  (f)  x'-+ ?/z  =  0. 

(c)  x"  -^  y'^-2zx  =  0.  (g)  xy  -Jryz  ^  zx  =  0. 

(d)  X  +  ?/  +  z  =  0.  (h)  x2  +  yz  +  xz  =  0. 

2.  Discuss  the  following  loci : 

(a)  x"  +  ?/2  =  z"  tan-  7.   (b)  y'^  -\-  z-  =  x-  tan^  a.    (c)  2-  +  x^  =  y^  tan^  /3. 

3.  Find  the  equation  of  the  cone  whose  vertex  is  the  origin  and  whose 
elements  cut  the  circle  x'^  ■]-  y'^  =  \Q,  z  =  2.  Ans.   x^  +  t/-  —  4z2  =  0. 

4.  A  point  is  equidistant  from  a  plane  and  a  line  perpendicular  to  the 
plane.    Prove  that  the  locus  is  a  cone. 

5.  A  point  moves  so  that  the  ratio  of  its  distances  from  two  lines  inter- 
secting at  right  angles  is  constant.  Prove  that  the  locus  is  a  cone.  What 
is  the  nature  of  the  locus  when  the  ratio  is  unity  ? 

6.  The  sum  of  the  distances  of  a  point  from  three  mutually  perpen- 
dicular planes  is  equal  to  its  distance  from  their  common  point  of  inter- 
section.   Show  that  the  locus  is  a  cone. 

117.  Surfaces  of  revolution.  The  surface  generated  by  revolv- 
ing a  curve  about  a  line  lying  in  its  plane  is  called  a  surface  of 
revolution. 

Familiar  examples  are  afforded  by  the  sphere,  and  the  right 
cvlinder  and  cone. 


SPECIAL  SURFACES 


306 


EXAMPLE 

Find  the  equation  of  the  surface  of  revolution  generated  by  revolv- 
ing the  ellipse  x^  +  4y^— 12x  =  0,  a  =  0,  about  the  X-axis. 

Solution.    Let  P  (x,  y,  z)  be  any  point  on  the  surface.    Pa.ss  a  plane 
through  P  and  OX  which  cuts  the  surface  along  one  position  of  the  ellipse, 
and  in  this  plane  draw  OY'  perpendicular  to  OX.  Referred  to  OX  and  OY' 
as  axes,  the  equation  of 
the  ellipse  is  evidently  ^^ 

(1)  x^+  'iy'^-12x  =  0. 

But  from  the  right  tri- 
angle PAB  we  get 

2/'2  =  2/2  +  2;2. 

Substituting  in  (1), 

(2)  x2  -f  4  2/2  +  4  22 

-12x  =  0. 

This  equation  expresses  the  relation  which  any  point  on  the  surface 
must  satisfy,  and  it  is  therefore  the  equation  of  the  surface. 

The  method  of  the  solution  enables  us  to  state  the 

Rule  to  Jirul  the  equation  of  the  surface  generated  by  revolving 

a  ctirve  in  one  of  the  coordinate  planes  about  one  of  the  axes  in 

that  plane. 

Substitute  in  the  equation  of  the  curve  the  square  root  of  the 

sum  of  the  squares  of  the  two  variables  not  measured  along  the 

axis  of  revolution  for  that  one  of  these   two   variables  which 

occurs  in  the  equation  of  the  curve. 

The  line  about  which  the  given  cui've  is  revolved  is  called  the 
axis  of  the  surface.  Sections  of  the  surface  by  planes  perpendicu- 
lar to  its  axis  are  obviously  circles  whose  centers  lie  on  the  axis. 

If  the  sections  of  a  surface  by  all  planes  perpendicular  to 
one  of  the  coordinate  axes  are  circles  whose  centers  lie  on  that 
axis,  then  the  surface  is  evidently  a  surface  of  revolution  whose 
axis  is  this  coordinate  axis.  This  enables  us  to  determine 
whether  or  not  a  given  surface  is  a  surface  of  revolution  whose 
axis  is  one  of  the  coordinate  axes. 


306  NEW  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Find  the  equations  of  the  surfaces  of  revolution  generated  by 
revolving  each  of  the  following  curves  about  the  axis  indicated,  and 
construct  the  figures: 

(a)  y'^  =  Ax  —  16,  X-axis.  Ans.   y"  +  z^=^  4x  —  16. 

(b)  x^  +  4 2/^  =  16,  I'-axis.  Ans.   x'^  +  -iy-  +  z-  =  16. 

(c)  x'^  =  4:Z,  Z-axis.  Ans.   x"  +  y"  =  4z. 

(d)  x2  -  2/2  =  16,  r-axis.  Ans.   x^-y^  +  z^  =  16. 

(e)  x'^  —  ?/2  =  16,  A'-axis.  Ans.   x^—  y^  —  z^  =  16. 

(f )  j^2  +  22  _  25,  Z-axis.  Ans.   x^  +  y^  +  z^  =  25. 

(g)  2/2  =  2pz,  Z-axis.    Ans.  A  paraboloid  of  revolution,  x^  +  y^  =  2pz. 

/j»2       '1/2  'j'2       oj2       2*2 

(h)  —-t-—  =  l,  A'-axis.    J.}is.  An  ellipsoid  of  revolution,  - -i-  —  -|-—=l. 

a-     b'^  a-     62     b- 

/J*  2         0/2 

(i) —  =  1,  r-axis. 

a"        ^'  o  2  2 

X  V  z 

Ans,  A  hyperboloid  of  revolution  of  one  sheet, \ —  =  1. 

9        Q  a-      6-      a- 

a^      b^  „       ,       , 

Ans.  A  hyperboloid  of  revolution  of  two  sheets,  -—— ;;  =  1- 

a^      62      52 

3.  Find  the  equations  of  the  surfaces  of  revolution  generated  by  revolv- 
ing each  of  the  following  curves  about  the  axis  indicated,  and  construct 
the  figures : 

(a)  x"  =  Az;  A'-axis.  (e)  xz  =  i  :  X-axis. 

(b)  ?/2  =  a;3;  X-axis.  (f)  xz  =  i;  Z-axis. 

(c)  x2  =  z  4-  4  ;  A^-axis.  (g)  y  =  x^  —  x  ;  X-axis. 

(d)  z2  —  x  —  3  ;  Z-axis.  (h)  z  =  sinx  ;  X'-axis. 

3.  Find  the  equation  of  and  construct  the  surface  formed  by  revolv- 
ing the  curve  z  =  c^  about  (a)  the  X-axis ;  (b)  the  Z-axis. 

4.  Verify  analytically  that  a  sphere  is  generated  by  revolving  a  circle 
about  a  diameter. 

5.  Find  the  equation  of  the  surface  of  revolution  genei-ated  by  revolv- 
ing the  circle  x^+  y"^—  2ax  +  a^—  r-=  0  about  the  F-axis.  Discuss  the 
surface  when  a>r,  a  =■  r,  and  a<r. 

Ans.  (x2  -I-  ?/2  +  22  4-  0-2  _  y2)2  _  4  (.^2  (j.2  ^  2.2).  Whcu  a>r  the  surface 
is  called  an  anchor  ring  or  torus. 

6.  Find  the  equations  of  the  cylinders  of  revolution  whose  axes  are 
the  coordinate  axes  and  whose  radii  equal  r. 

Ans.   y-  +  z'^  =  r^ ;  z^  +  x'^  =  r"^ ;  x^  +  y"^  =  r'^. 


SPECIAL  SURFACES  307 

7.  Find  the  equations  of  the  cones  of  revolution  whose  axes  are  the 
coordinate  axes  and  wliose  elements  make  an  angle  of  0  with  the  axis  of 
revolution.    Ans.    y'^+ z'^  =  xHan^(p;  z'^+ x'^=y'^tan^<p;  x^+y'^=z^ta,n^<(>. 

8.  Show  that  the  following  loci  are  surfaces  of  revolution  : 

(a)  2/2  +  z2  =  4x.  (f)  (x2+22)2/:^4a2{2a-?/). 

(b)  x2-4?/2+22=  0.  (g)  x^  +  y^+zx^+zy^-z  +  3  =  0. 

(c)  4x2  +  4  2/2  _  ^2  ^  16.  (h)  X*  -y*  +  z*+2  x'^z"-  =  1. 

(d)  x2-4?/2+22_3y  =  0.  (i)    X2+  i/2+z3_2y  +  1  =0. 

(e)  X22  +  X2/2  —  3. 

9.  A  point  moves  so  that  its  distance  from  a  fixed  plane  is  in  a  con- 
stant ratio  to  its  distance  from  a  fixed  point.  Show  that  the  locus  is  a 
surface  of  revolution. 

10.  A  point  moves  so  that  Its  distance  from  a  fixed  line  is  in  a  constant 
ratio  to  its  distance  from  a  fixed  point  on  that  line.  Prove  analytically  that 
the  locus  is  a  cone  of  revolution.   What  values  of  the  ratio  are  excluded  ? 

118.  Ruled  surfaces.  A  surface  generated  by  a  moving  straight 
line  is  called  a  ruled  surface.  If  the  equations  of  a  straight  line 
involve  an  arbitrary  constant,  then  the  equations  represent  a 
system  of  lines  which  form  a  ruled  surface.  If  we  eliminate 
the  parameter  from  the  equations  of  the  line,  the  result  will  be 
the  equation  of  the  ruled  surface. 

For  if  {pc^,  7/j,  z^  satisfy  the  given  equations  for  some  value 
of  the  parameter,  they  will  satisfy  the  equation  obtained  by 
eliminating  the  parameter;  that  is,  the  coordinates  of  every 
point  on  every  line  of  the  system  satisfy  that  equation. 

Cylinders  and  cones  are  the  simplest  ruled  surfaces. 

EXAMPLES 

1.  Find  the   equation  of  the  surface  generated  by  the  line  whose 

equations  are  ,  1 

x  +  y  —  kz,x  —  y=-z. 
k 

Solution.    We  may  eliminate  k  from  these  equations  of  the  line  by 

multiplying  them.    This  gives 

(1)  x2-?/2=:22_ 

This  is  the  equation  of  a  cone  (Art.  116)  whose  vertex  is  the  origin.  As 
the  sections  made  by  the  planes  x  =  k  are  circles,  it  is  a  cone  of  revolu- 
tion whose  axis  is  the  A'-axis. 


308 


NEW  ANALYTIC   GEOMETRY 


We  may  verify  that  the  given  line  lies  on  the  surface  (1)  for  all  values 
of  /c  as  follows  : 

Solving  the  equations  of  the  line  for  x  and  y  in  terms  of  z,  we  get 

x  =  -{k  +  -\z,     y  =  -[k )z. 

2\        kj  2\        k) 

Substituting  in  (1), 


^(-r^'4('-*/ 


an  equation  which  is  true  for  all  values  of  k  and  z,  as  is  seen  by  removing 
the  parentheses.  Hence  every  point  on  any  line  of  the  system  lies  on  (1), 
since  its  coordinates  satisfy  (1). 

2.  Determine  the  nature  of  the  surface  z^  —  3zx  4-  8?/  =  0. 

Solution.   The  intersec- 
tion of  the  surface  with  the 
plane  z  =  k  is  the  straight 
line 
A-3-3A:a;  +  8?/  =  0,     z  =  k. 

Hence  the  surface  is  the 
ruled  surface  generated  by 
this  line  as  k  varies.  To 
construct  the  surface  con- 
sider the  intersections  with 
the  planes  x  ==  0  and  x  =  8. 
Their  equations  are  respec- 

ti^^ly  X  =  0, 

8  (/  +  ^3  =  0  ; 

and  X  =  8, 

By-  24  z  +  23  =  0. 

Joining  the  points  on  these  curves  which  have  the  same  value  of  z  gives 
the  lines  generating  the  surface. 

The  method  used  in  Ex.  2  is  adapted  to  the  determination 
and  construction  of  ruled  surfaces.  An  examination  of  the 
equation  of  such  a  surface  will  suggest  a  system  of  planes 
whose  intersections  with  the  surface  are  a  system  of  lines,  as 
illustrated  in  Problem  2  on  the  following  page. 


SPECIAL  SURFACES  309 

PROBLEMS 

1.  Show  that  the  following  loci  are  ruled  surfaces  whose  generators  are 
parallel  to  one  of  the  coordinate  planes.    Construct  and  discuss  the  loci : 

(a)  z  —  xy  =  0.  (f )  j/2  _  x^z. 

(b)  x^y  -  z2  =  0.  (g)  y  =  xz{2-  z)^. 

(c)  z'^  -  zx  +  y  =  0.  (h)  2/2  _  a.2(22  + 1^ 

(d)  xhj  +  xz  =  y.  (i)  2/2  =  x2{z2-l). 
{e)y-xz"'  =  0.  (j)  2/2^x2(1 -22). 

Remark.  The  surfaces  may  be  easily  constructed  from  string  and  cardboard. 

2.  Show  that  the  following  loci  are  ruled  surfaces  : 

(a)  (x  +  2/)z+  (x  +  ?/)2-l  =  0. 

(b)  j;2  _  2  a-2  -  ?/2  +  22  =  3. 

(c)  2/2  +  4  ^2  ^  jy  _  4  ^2  _  2  xz  +  3  =  0. 

(d)  X3  +  3  ?/x2  -  X22  _  3  2/22  _  .^2  ^  ^2  _  Q. 

(e)  x2  -  2/2  =  2. 

(f)  X2  -  2/2  =22-1. 

/fi/i^  Find  a  system  of  planes  which  cut  the  surface  in  a  system  of  straight 
lines. 

3.  Find  the  equations  of  the  ruled  surfaces  whose  generators  are  the 
following  systems  of  lines,  and  discuss  the  surfaces  : 

(a)  X  +  y  =  k,  k  {x  —  y)  =  a?.  Ans.  x^  —  y^  =  a^. 

(h)  'ix  —  2y  =  kz,k{4:X  + 2y)  =z.  Ans.  16x^  —  4y^  =  z-. 

(c)  X  —  2  y  —  4  kz,  k  {x  —  2  y)  =  4.  Ans.  x^  —  4  2/2  =  16  2. 

(d)  X  +  ky  +  4  z  =  4  k,  kx  —  y  —  4  kz  =  4.  Ans.  x2  +  2/2  —  16  22  =  16. 

(e)  X  —  y  —  kz  =  0,  X  —  z  —  ky  =:  0. 

(f )  Sx-  z-k  =  0,  ky  -z  =  0. 

4.  Given  two  planes,  one  with  a  variable  intercept  on  the  A'-axis,  the 
other  with  a  variable  intercept  on  the  Y-axis.  The  remaining  intercepts 
being  unity,  find  the  equation  of  the  ruled  surface  generated  by  the 
line  of  intersection  of  these  planes 

(a)  when  their  variable  intercepts  are  in  the  ratio  1  : 2. 

(b)  when  their  distances  from  the  origin  are  in  the  ratio  1  : 3. 

Ans.    [y  (z  +  7/)f  -  [3x  (2  +  x)]^  =  {4xy)^. 

(c)  when  the  sum  of  their  distances  from  the  origin  is  unity. 


CHAPTER  XVIII 


TRANSFORMATION  OF  COORDINATES.    DIFFERENT  SYSTEMS 
OF  COORDINATES 

119.  Translation  of  the  axes.  Formulas  applicable  to  space, 
entirely  analogous  to  those  established  in  Chapter  IX  for  the 
plane,  are  derived  as   ex- 


Z'< 


z\- 


(h,k,/ 


±, 


/X 


plained  below. 

Theorem.  17ie  equations 
for  translating  the  axes  to 
a  neiv  origin  O'  (h,  k,  V)  are 

(I)  x  =  x^  +  h, 

y  =  y'  +  k, 

z  =  z'  -{-1. 

Proof.  Let  the  coordi- 
nates of  any  point  before 

and  after  the  translation  of  the  axes  be  (x,  y,  z)  and  {x\  y\  «') 
respectively.  Projecting  OP  and  OO^P  on  each  of  the  axes,  we 
get  equations  (I).  q.e.d. 

120.  Rotation  of  the  axes.  Simple  formulas  for  rotation  arise 
if  two  of  the  axes  are  rotated  about  the  third.  For  example, 
when  the  axes  OX  and  OY  are  turned  through  an  angle  0  about 
the  Z-axis,  the  ^.'-coordinate  of  any  point  P  does  not  change, 
and  the  new  x-  and  ?/-eoordinates  are  given  by  formulas  (II), 
Art.  55.    Hence  the 

Theorem.  The  equations  for  rotating  the  axes  about  the  Z-axis 
through  an  angle  6  are 

(II)  jr=  jr'costf  —  y'sin^,     y  =  x^  s\nd -\- y' co&d ,     z  =  z\ 

310 


TRANSFORMATION  OF  COORDINATES 


311 


Similar  formulas  result  when  the  axes  are  rotated  about  OY 
or  OX. 

If  the  axes  are  rotated  about  the  origin  into  the  new- 
position  0-X'Y'Z',  and  if  the  coordinates  of  any  point  P 
before  and  after  the  rotation  are 
respectively  (x,  y,z)  and  (x\  y\  z'), 
we  have  the 

Theorem.  Jf  a^,  (3^,  y^;  a^,  fS,^,  y^- 
and  a^,  /S^,  y.^,  are  resj^ectiveli/  the 
direction  angles  of  the  three  mtitu- 
ally  perpendicular  lines  OX',  OY', 
and  OZ',  then  the  equations  for 
rotating  the  axes  to  the  position 
O-X'Y'Z'  are 


(III) 


X  =  x'  cosa^  +  z/'cosa2  +  z'  cosa^. 


i  y  =  x'cos/3j^  -{-y'cosp^  +  z'cos^3, 
}^z  =  jr'cosy^  + i/'cosy2  +  ^'cosy3.* 

Proof  Projecting  OP  and  OA'B'P  on  each  of  the  axes  OX, 
OY,  and  OZ,  we  obtain  immediately  equations  (III).        q. e.d. 

Theorem.  The  degree  of  an  equation  is  unchanged  by  a  trans- 
forination  of  coordinates. 

This  may  be  shown  by  reasoning  as  in  Art.  57. 


PROBLEMS 

1.  Transform  the  equation  x'^-\-y'^  —  Ax-\-2y  —  Az  +  l^Qhj  trans- 
lating the  origin  to  the  point  (2,  —  1,  —  1).  Ans.   x^  +  2/^  —  4 2  =  0. 

2.  Derive  the  equations  for  rotating  the  axes  through  an  angle  6  about 
(a)  tlie  X-axis ;  (b)  the  Y-axis. 

*  The  direction  cosines  of  OX',  OY',  and   OZ'  obviously  satisfy  the  six 
equations 
cos-  at  +  cos2  /3i  +  cos2  71  =  1,    cos  a\  cos  a.i  +  cos  ^i  cos  ^2  +  cos  7^  cos  72  =  0, 
cos2  ai  +  cos^  ^2  +  cos2  72  ==  1,    cos  a^  cos  a^  +  cos  /Sg  cos  ^^  +  cos  72  cos  73  =  0, 
cos^  u'g  +  cos2  /33  +  cos2  73  =  1 ,    cos  a^  cos  a  I  +  cos  /Sg  cos  /3i  +  cos  73  cos  7i  =  0. 

Hence  only  three  of  the  nine  constants  in  (III)  are  independent. 


312  NEW  AXALYTIC   GEOxMETRY 

3.  Show  that  the  following  equations  may  be  transformed  into  the 
given  answers  by  translating  the  axes,  or  by  rotating  them  about  one  of 
the  coordinate  axes  (see  Art.  71)  : 

(a)  X-  +  y-^  -  Z'  -  6x  -  8y  +  lOz  =  0.  Ans.   x-  +  y- -  z"^  =  0. 

(b)  3x2  — 8x2/  + 32/2— 522+ 5  =  0.  -^ns.   x''- -  1  y^  +  bz^  =  b. 

(c)  2/2  +  4z2  -  16x  -  6y  +  16z  +  9  =  0.  Ans.   y^  +  4z2  =  \Qx. 

(d)  2x2-52/2- 522- 62/3  =  0.  Ans.   x^-4y^-z-  =  0. 

(e)  9x2  _  252/2  +  16z2  -  242X  -  80x  -  GOz  =  0.    Ans.   x^-y^  =  Az. 

4.  Show  that  Ax  +  By  +  Cz  +  D  =  0  may  be  reduced  to  the  form  x  =  0 
by  a  transformation  of  coordinates. 

Hint.  Remove  the  constant  term  by  translating  the  axes,  then  remove  the 
2-teim  by  rotating  the  axes  about  the  F-axis,  and  finally  remove  the  y-term 
by  rotating  about  the  Z-axis. 

5.  Transform  the  equation  5  x2  -|-  8  2/'^  +  522  —  4  2/z  +  8  2X  +  4 x?/  —  4  x  + 
2?/  +  4z  =  0  by  rotating  the  axes  to  a  position  in  which  their  direction 
cosines  are  respectively  f,  f,  i  ;  i,  —  I,  f  ;  J,  —  i,  —  I- 

Ans.    3  x2  4-  8  2/2  =  2  2. 

6.  Show  that  the  xy-term  may  always  be  removed  from  the  equation 
.4x2  +  2)2/2  +  C22  +  Fxy  +  ii  =  0  by  a  rotation  about  the  Z-axis. 

7.  Show  that  the  2/2-term  may  always  be  removed  from  the  equation 
^x2  +  By^  +  Cz^  +  Dyz  +  ^  =  0  by  rotating  about  the  X-axis. 

8.  What  are  the  direction  cosines  of  OX,  OF,  and  OZ  (Fig.,  p.  311) 
referred  to  0X\  OY',  and  OZ'  ?    What  six  equations  do  they  satisfy  ? 

9.  Show  that  the  six  equations  obtained  in  Problem  8  are  equivalent 
to  the  six  equations  in  the  footnote,  p.  311. 

10.  If  (x,  y,  2)  and  (x',  y',  z')  are  respectively  the  coordinates  of  a 
point  before  and  after  a  rotation  of  the  axes,  show  that 

X'  -\-  y"  +  Z^  =  X'2  +  ?/'2  -I-  2'2_ 

11.  The  possibilities  of  simplifying  an  equation  by  rotation  of  the  axes 
will  be  stated  here  without  proof.  Consider  the  equation  of  the  second 
degree 

Ax"  +  By-  +  Cz'  4-  Byz  +  -E'2X  ^-  Fxy  -^^  Gx  +  Hy  ^  Iz  ^  K  =  0. 
It  is  shown  in  more  advanced  treatises  that  it  is  always  possible  to  de- 
termine at  least  one  new  system  of  rectangular  axes  OA'',  01"',  and  OZ', 
having  the  same  origin  as  the  old  axes,  and  .such  that  the  tran.sformed 
equation  with  reference  to  these  new  axes  will  lack  terms  in  xy,  yz,  and  zx. 
This  simplification  of  the  equation  is  accomplished  by  a  transformation  of 
the  form  (III),  that  is,  a  rotation  of  the  axes  about  the  origin. 


TRANSFORMATION  OF  COORDINATES 


313 


121.  Polar  coordinates.  The  line  OP  drawn  from  the  origin 
to  any  point  P  is  called  the  radius  vector  of  P.  Any  point 
P  determines  four  numbers,  its 
radius  vector  p,  and  the  direction 
angles  of  OP,  namely  a,  p,  and  y, 
which  are  called  the  polar  coordi- 
nates of  P. 

These  numbers  are  not  all  independ- 
ent, since  a,  /3,  and  7  satir.fy  (II),  Art.  88. 
If  two  are  known,  the  third  may  then  be 
found,  but  all  three  are  retained  for  the 
sake  of  symmetry.  / 

Conversely,  any  set  of  values 
of  p,  a,  /?,   and   y  which   satisfy   (II),  Art.    88,   determine   a 
point  whose  polar  coordinates  are  p,  a,  (3,  and  y. 

Projecting  OP  on  each  of  the  axes,  we  get  the 

Theorem.    The  equations  of  transformation  from  rectangiilar 
to  polar  coordinates  are 
(IV)  x  =  pcosa,     y  =  pcosp,     z=y9cosy. 

Obviously 
(1)  p'  =  x'  +  y-'  +  z\ 

which  expresses  the  radius  vector  in  terms  of  x,  y,  and  z. 

122.  Spherical  coordinates.  Any  point 
P  determines  three  numbers,  namely, 
its  radius  vector  p,  the  angle  6  be- 
tween the  radius  vector  and  the  Z-axis, 
and  the  angle  <^  between  the  projection 
of  its  radius  vector  on  the  A'F-plane 
and  the  A-axis.  These  numbers  are 
(tailed  the  spherical  coordinates  of  P. 
6  is  called  the  colatitude  and  <f>  the 
longitude. 

Conversely,  given  values  of  p,  6,  and  <^  determine  a  j)oint 
P  whose  spherical  coordinates  are  (p,  6,  4>)- 


314 


NEW  ANALYTIC   GEOMETRY 


Projecting  OP  on  OA ,     OM  =  p  sin  6, 
and  projecting  OP  and  OMP  on.  eacli  of  the  axes,  we  prove  the 

Theorem.  The  equations  of  transformation  from  rectangular 
to  spherical  coordinates  are 

(V)  x  =  ps,\ndcos<f>,     y  =  p  sin  0  sin  ^,     z  z=  p  cos  6. 

The  equations  of  transformation  from  splierical  to  rectangular 
coordinates  may  be  obtained  by  solving  (V)  for  p,  6,  and  <f>. 

123.  Cylindrical  coordinates.  Any  point  P  (x,  ?/,  z)  determines 
three  numbers,  its  distance  ?:  from  the  A'F-plane  and  the  polar 
coordinates  (r,  <f>)  of  its  projection  (x,u,0)  on  the  A'F-plane. 
These  three  numbers  are  called  the  cylindrical  coordinates  of  P. 
Conversely,  given  values  of  r,  </>,  and  z  de-  ^.^ 

termine  a  point  whose  cylindrical  coordi- 
nates are  (r,  ^,  z).  Then  we  have  at  once  the 

Theorem.  Tlie  equations  of  transforma- 
tion from  rectanrjular  to  cylindrical  coordi- 
nates are 

(VI)  j:=rcos^,   z/=rsin^,  z  =  z. 

The  equations  of  transformation  from  cylindrical  to  rectangu- 
lar coordinates  may  be  obtained  by  solving  (VI)  for  ?•,  «^,  and  z. 


PROBLEMS 

1.  AVhat  is  meant  by  the  "locus  of  an  equation"  in  the  polar  coordi- 
nates /D,  cr,  /3,  and  7?  in  the  spherical  coordinates  p,  (9,  and  0?  in  the 
cylindrical  coordinates  r,  0,  and  z  ? 

2.  How  may  the  intercepts  of  a  surface  on  the  rectangular  axes  be  found 
if  its  equation  in  polar  coordinates  is  given?  if  its  eciuation  in  spherical 
coordinates  is  given?  if  its  equation  in  cylindrical  coordinates  is  given? 

3.  Transform  the  following  equations  into  polar  coordinates  : 

(a)  z"  4-  v"  +  z^  =  25.  Ans.    p  =  5. 

(b)  x^  +  y-  —  z~  =  0.  Ans.    7  =  -  • 

(c)  2  x2  —  !/■- —  22  =  0.  -4ns.    a  =  cos-i^V3. 


TRANSFORMATION  OF  COORDINATES  SIS 

4.  Transform  the  following  equations  into  spherical  coordinates  : 
(a.)  x^  +  y^  +  z"  =  16.  Ans.    p  =  i. 

(h)  2x  +  Sy  =  0.  Ans.    0  =  tan- 1  (- |). 

(c)  Sx^+Sy-  =  7z\  Ans.    ^  =  tan-iiV21. 

6.  Transform  the  following  equations  into  cylindrical  coordinates: 

(a)  5x  —  y  =  0.  Ans.    (^  =  tan-i  5. 

(b)  a;2  +  y2.  -  4,  ^^s_    ,.  _  2. 

6.  Find  the  equation  in  polar  coordinates  of 

(a)  a  sphere  whose  center  is  the  pole. 

(b)  a  cone  of  revolution  whose  axis  is  one  of  the  coordinate  axes. 
Ans.    (a)  p  =  constant;  (b)  a  =  constant,  /3  =  constant,  or  7  =  constant. 

7.  Find  the  equation  in  spherical  coordinates  of 

(a)  a  sphere  whose  center  is  the  origin. 

(b)  a  plane  through  the  Z-axis. 

(c)  a  cone  of  revolution  whose  axis  is  the  Z-axis. 

Ans.    (a)  p  =  constant ;  (b)  <;&  =  constant ;  (c)  0  —  constant. 

8.  Find  the  equation  in  cylindrical  coordinates  of 

(a)  a  plane  parallel  to  the  A'F-plane. 

(b)  a  plane  through  the  Z-axis. 

(c)  a  cylinder  of  revolution  whose  axis  is  the  Z-axis. 

Ans.    (a)  z  =  constant;  (b)  0  =  con!?tant;  (c)  ?■  =  constant. 

9.  In  rectangular  coordinates  a  point  is  determined  as  the  intersec- 
tion of  three  mutually  perpendicular  planes.    Show  that 

(a)  in  polar  coordinates  a  point  is  regarded  as  the  intersection  of  a 
sphere  and  three  cones  of  revolution  which  have  an  element  in  common. 

(b)  in  spherical  coordinates  a  point  is  regarded  as  the  intersection  of  a 
sphere,  a  plane,  and  a  cone  of  revolution  which  are  mutually  orthogonal. 

(c)  in  cylindrical  coordinates  a  point  is  regarded  as  the  intersection 
of  two  planes  and  a  cylinder  of  revolution  which  are  mutually  orthogonal. 

10.  Show  that  the  square  of  the  distance  r  between  two  points  whose 
polar  coordinates  are  (pj,  a^,  /3^,  y^)  and  (p.,,  tr.,,  /Sj,  7")  is 

j'2  =  p~  4.  p I  _  2  p,  P2  (cos  <a^i  cos  a^  +  cos/3^  cos^„  -1-  cos7j  cos7„). 

11.  Find  the  general  equation  of  a  plane  in  polar  coordinates. 

Ans.   p  {A  cos  a  +  B  cos  /S  +  C  cos  7)  -|-  Z>  =  0. 

12.  Find  the  general  equation  of  a  sphere  in  polar  coordinates. 

Ans.   p^  +  p{G  cos  a  +  H  cos ^  +  1  cos 7)  4-  K  =  0. 


CHAPTER  XIX 

QUADRIC  SURFACES  AND  EQUATIONS  OF  THE  SECOND 
DEGREE  IN  THREE  VARIABLES 

124.  Quadric  surfaces.  The  locus  of  an  equation  of  the  sec- 
ond degree  in  x,  y,  and  z,  of  which  the  most  general  form  is 

(1)   ^ x^-\- Bif+  Cz'-^  Dijz  +  Ezx  +  Fxy -\-Gx+Hy-\- Iz  +  A'  =  0, 

is  called  a  quadric  surface  or  conicoid.  We  may  learn  something 
of  the  nature  of  such  a  surface  by  taking  cross  seetioas.  We 
first  obtain 

Theorem  I.  The  intersection  of  a  quadric  with  any  plane  is  a 
conic  or  a  degenerate  conic. 

Proof.  By  a  transformation  of  coordinates  any  plane  may  be 
made  the  A'F-plane,  z  =  0.  Referred  to  any  axes  the  equation 
of  a  quadric  has  the  form  (1)  (Theorem,  p.  311).  Hence  the 
equation  of  the  curve  of  intersection  referred  to  axes  in  its 
own  plane  5;  =  0  is 

A  x^  +  Fxy  +  B7f  +  Gx  -\-Hy-{-K  =  0, 

and  the  locus  is  therefore  a  conic  or  a  degenerate  conic,  by 
Art.  70.  Q.E.D. 

As  already  pointed  out  in  Art.  71,  the  parabola,  ellipse,  and 
hyperbola  were  originally  studied  as  conic  sections, — plane  sec- 
tions of  a  conical  surface.  From  the  preceding  theorem  and 
by  intuition,  the  truth  of  the  following  statement  is  manifest. 

Corollary.  The  curve  of  intersection  of  a  cone  of  revolution 
lo'ith  a  plane  is  an  ellipse,  hyperhola,  or  parabola,  accordlncj  as 
the  2:)lane  cuts  all  of  the  elements,  is  parallel  to  two  elements 

316 


QUADRIC   SURFACES  317 

(cutting  the  other  elements  —  some  on  one  side  of  the  vertex 
and  some  on  the  other'),  or  is  paixdlel  to  one  element  (cutting 
all  the  others  on  the  same  side  of  the  vertex). 

For  sections  of  a  quadric  by  a  set  of  parallel  planes,  the 
following  result  is  important : 

Theorem  II.  The  sections  of  a  quadric  with  a  system  of  paral- 
lel planes  are  conies  of  the  same  species. 

The  truth  of  this  statement  is  established  in  the  following 
sections.  Tiie  meaning  of  the  theorem  is  this  :  A  set  of  parallel 
sections  will  all  be  ellipses,  or  all  hyperbolas,  or  all  parabolas, 
the  exceptional  cases  (Art.  70)  under  each  species  being  included. 

125.  Simplification  of  the  general  equation  of  the  second  degree 
in  three  variables.  If  equation  (1)  be  transformed  by  rotating 
the  axes,  it  can  be  shown  that  the  new  axes  may  be  so  chosen 
that  the  terms  in  yz,  zx,  and  xy  will  drop  out  (Problem  11, 
p.  312).    Hence  (1)  reduces  to  the  form 

A  'x"^  +  i3y  +  C'z'^  +  G'x  +  iVy  +  I'z  +  A''  =  0. 

Transforming  this  equation  by  translating  the  axes,  it  is  easy 
to  show  that  the  axes  may  be  so  chosen  that  the  transformed 
equation  will  have  one  of  the  two  forms 

(1)  A  "x'  +  B'Y  +  C"z-  +  K"  =  0, 

(2)  A"x^ -\- B'Y -\- ^"^  =  0. 

Note  the  difference  in  (1)  and  (2).  In  (1)  all  the  squares  and 
no  first  powers  are  represented,  in  (2)  only  two  squares  and 
the  hrst  power  of  the  other  variable. 

If  all  of  the  coefficients  in  (1)  and  (2)  are  different  from 
zero,  they  may,  with  a  change  in  notation,  be  respectively  writ- 
ten in  the  forms 

(3)  ±rU|-!±5'  =  -'- 


(4)  !-.±7i  =  2... 


a^^  b^ 
I 


318  NEW  ANALYTIC  GEOMETllY 

The  purpose  of  the  following  sections  is  to  discuss  the  loci  of 
these  equations,  which  are  called  central  and  noncentral  quadrics 
respectively. 

If  one  or  more  of  the  coefficients  in  (1)  or  (2)  are  zero,  the 
locus  is  called  a  degenerate  quadric. 

Certain  cases  are  readily  disposed  of  by  means  of  former 
results. 

If  /v"  =  0,  the  locus  of  (1)  is  a  cone  (Theorem,  Art.  IIG) 
unless  the  signs  of  A",  B",  and  C"  are  the  same,  in  which  case 
the  locus  is  z,XJomt,  namely  the  origin. 

If  one  of  the  coefficients  A",  B",  and  C"  is  zero,  the  locus  is  a 
eijUnder  whose  elements  are  parallel  to  one  of  the  axes  and  whose 
directrix  is  a  conic  of  the  elliptic  or  hyperbolic  type.  If  also 
A'"  =  0,  the  locus  will  be  a  pair  of  intersecting  jAanes. 

If  two  of  the  coefficients  A  ",  B",  and  C"  are  zero,  the  locus  is  a 
pair  of  parallel  jolanes  (coincident  if  7v"  =  0),  or  there  is  no  locus. 

If  one  of  the  coefficients  in  (2)  is  zero,  the  locus  is  a  cylinder 
whose  directrix  is  a  parabola,  or  di,  pair  of  intersecting  p)lanes. 

If  tioo  of  the  coefficients  are  zero,  the  locus  is  ajmir  of  coinci- 
dent planes.  (A"  and  B"  cannot  be  zero  simultaneously,  as  the 
equation  would  cease  to  be  of  the  second  degree.) 

PROB)  EMS 

1.  Construct  and  discuss  the  loci  of  the  following  equations: 

(a)  9x^-362/2+ 422  =  0.  (e)  4y^-25  =  0. 

(b)  16x2- 4 2/2 _  22  =  0.  (f)  3  2/2  +  702  =  0. 

(c)  4x2  + z2_  10  =  0.  (g)  8?/2+25z  =  0. 

(d)  1/2  -  9z2  +  36  =  0.  (h)  z2  +  16  =  o. 

2.  Show  by  transformation  of  coordinates  that  the  following  quadrics 
are  degenerate : 

(a)  X2  -  ?/  +  22  _  0  2  +  9  =  0. 

(b)  x2  +  4  2/2  _  z2  _  2  x  +  8  2/  +  5  =  0. 

(c)  x2  +  2/2  +  z2  +  2  X  -  2  2/ +  4  z  +  6  =  0. 

(d)  x2  +  2/2  •  -  2  22  +  2  2/  +  4  2  -  1  =^  0. 

(e)  x2  +  2/z  =  0. 


QUADRIC  SURFACES 


319 


A"  x^    ^     ^ 

126.  The  ellipsoid 1 1 =1.   If  all  of  the  coefficients  in 

<r      \r      & 

(3),  Art.  125,  are  positive,  the  locus  is  called  an  ellipsoid.    A  dis- 
cussion of  its  equation  gives  us  the  following  properties : 

1.  The  ellipsoid  is  symmetrical  with  respect  to  each  of  the 
coordinate  planes  and  axes  and  the  origin.  These  planes  of 
symmetry  are  called  the  principal  planes  of  the  ellipsoid. 

2.  Its  intercepts  on  the  axes  are  respectively 

x^±a,      y=±h,      z=±c. 

The  lines  AA'  =  2  a,  BB'  =  2  b,  CC  =  2  c,  are  called  the 
axes  of  the  ellipsoid  (see  figure  below). 

3.  Its  traces  on  the  principal  planes  are  the  ellipses  ABA'B', 
BCB'C,  and  ACA'C',  whose  equations  are 

^  4-  ^'  =  1         y!  4.  -  -  1         ^'  -L  ?-'  -  1 

a''^  b'         '      b'^  c'         '     a-       c'~ 

4.  The  equation  of  the  curve  in  Avhich  a  j^lane  parallel  to  the 
A'F-plane,  z  —  k,  intersects  the  ellipsoid  is 

^2  .,2  J.2 


(1)       V2  + 


'-7-' 


or 


+ 


?/ 


=  1. 


If  k  increases 


c^  ^  ^       c 

Tlie  locus  of  this  equation  is  an  ellipse 
iiom  0  to  c,  or  de- 
creases from  0  to 
—  c,  the  plane  recedes 
from  the  A"F-plane, 
and  the  axes  of  the 
ellipse  decrease  from 
2  a  and  2  b  respec- 
tively to  0,  when 
the  ellipse  degener- 
ates into  a  point.  If 
A; > c  or  k^—  r,  there  is  no  locus.  Hence  the  ellipsoid  lies 
entirely  between  the  planes  z  =±c. 


320 


NEW  ANALYTIC  GEOMETRY 


111  like  manner  tlie  sections  parallel  to  the  FZ-plane  and  the 
ZA'-plane  are  ellipses  whose  axis  decrease  as  the  planes  recede. 
Hence  the  ellipsoid  lies  entirely  between  the  planes  x  =  ±_a 
and  y  ^  -^h.    The  ellipsoid  is  therefore  a  closed  surface. 

If  a  =  b,  the  section  (1)  is  a  circle  for  values  of  k  such  that 
—  c  <  J:  <  c,  and  hence  the  ellipsoid  is  now  an  ellipsoid  of 
revolution  whose  axis  is  the -Z-axis.  If  ^>  =  c  or  c  =  a,  it  is  an 
ellipsoid  of  revolution  whose  axis  is  the  A'-  or  I'-axis. 

If  a.  =  i  =  c,  the  ellipsoid  is  a  sphere,  for  its  equation  may 

be  written  in  the  form  x^  +  y^  +  ^"  =  o'^- 

x^       ip'       z^ 
127.  The  hyperboloid  of  one  sheet 1 =  1.    If  two  of 

q2         ^         ^ 

the  coefficients  in  (3),  Art.  125,  are  positive  and  one  is  negative, 
the  locus  is  called  a  hyperboloid  of  one  sheet.  Consider  first  the 
equation 

(1) 


a-       h~       c^ 


^2 
~2 


A  discussion  of  this  equation 
gives  us  the  following  properties  : 

1.  The  hyperboloid  is  symmetri- 
cal with  respect  to  each  of  the  coordi- 
nate planes  and  axes  and  the  origin. 

2.  Its  intercepts  on  the  X-axis 
and  the  F-axis  are  respectively 

•'■  =  ±  <h     y  =  ±  ^, 
but  it  does  not  meet  the  Z-axis. 

3.  Its  traces  on  the  coordinate 
planes  are  the  conies 

of  which  the  first  is  the  ellipse  whose  axes  are  A  A'  =  2  a  and 
BB'  —2h,  and  the  others  are  the  hyperbolas  whose  transverse 
axes  are  BB'  and  AA '  respectively. 


^2  y2 

a'  "^  b'' 


m    ^^I^^BfU^P 

o       o* 


QUADRIC  SURFACES  321 

4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 

J^F-plane,  z  =  k,  intersects  the  hyperboloid  is 

(2)        -2+r2=l  +  T2'     or +         -^ 


^(c^+F)      -A<^'  +  k') 

c-  ^       c  ' 

The  locus  of  this  equation  is  an  ellipse.  If  k  increases  from 
0  to  CO  ,  or  decreases  from  0  to  —  00  ,  the  plane  recedes  from  the 
AT-plane,  and  the  axes  of  the  ellipse  increase  indefinitely  from 
2  a  and  2  h  respectively.  Hence  the  surface  recedes  indefiniteh' 
from  the  X  F-plane  and  from  the  Z-axis. 

In  like  manner  the  sections  formed  b)^  the  planes  x  =  k'  and 
y  =  A;"  are  seen  to  be  hyperbolas.  As  /:'  and  A;"  increase  numer- 
ically, the  axes  of  the  hyperbolas  decrease,  and  when  /.'=  ±  a 
or  A;"  =  ±_  h,  the  hyperbolas  degenerate  into  intersecting  lines. 
As  k'  and  k"  increase  beyond  this  point,  the  directions  of  the 
transverse  and  conjugate  axes  are  interchanged,  and  the  lengths 
of  these  axes  increase  indefinitely. 

The  hyperboloid  (1)  is  said  to  "  lie  along  the  Z-axis." 

The  equations 

x^      iF      z^  x^      iF      ^ 

(<3)  —^  TjH        ^=^l>  2'72~'        2^^-^J 

^  ^  a-       b-       (•'■  cr       b^       r 

are  the  equations  of  hyperboloids  of  one  sheet  which  lie  along 
the  I'-axis  and  the  ^-axis  respectively. 

li  a  =  b,  the  hyperboloid  (1)  is  a  surface  of  revolution  whose 
axis  is  the  Z-axis,  because  the  section  (2)  becomes  a  circle. 
The  hyperboloids  (3)  will  be  hyperboloids  of  revolution  if  a  =  c 
and  b  =  c  respectively. 

X^  y2  2^2 

128.  The  hyperboloid  of  two  sheets =  1.    If  only 

a^      b^      (^ 

one;  of  the  coefficients  in  (3),  Art.  125,  is  positive,  the  locus  is 
called  a  hyperboloid  of  two  sheets.    Consider  first  the  equation 

iC  ?/  Z' 

^  ■'  a^      V      c^ 


322 


NEAV  ANALYTIC   GEOMETRY 


1.  The  hyperboloid  is  symmetrical  with  respect  to  each  of 
the  coordinate  planes  and  axes  and  the  origin. 

2.  Its  intercepts  on  the  A'-axis  are  x  =  ±  a,  but  it  does  not 
cut  the  F-axis  and  the  Z-axis. 

3.  Its  traces  on  the  A'F-plane  and  the  AZ-plane  are  respec- 
tively the  hyperbolas 


--  =  1 

..2  -^' 


^  ~  i^ "    '     a^      r 
which  have  the  same  transverse  axis  AA'  =  2a,  but  it  does 
not  cut  the   FZ-plane. 

4.  The  equation  of  'the  curve  in  which  a  plane  parallel  to 
the  FZ-plane,  x  =  k,  intersects  the  hyperboloid  (1)  is 
Z.-2  >fi  -y^ 


V'^      z-      k'     _ 
—  -\ —  = 1 


or 


a 


■  + 


r.2 


=   1. 


«■")  -2(^'-«') 


This  equation  has  no  locus  if  —  a  <  k<a.  If  k  =  ±  a,  the 
locus  is  a  point  ellipse,  and  as  k  increases  from  a  to  oo,  or 
decreases  from  —a  to  —  cc,  the  locus  is  an  ellipse  whose 
axes  increase  indefinitely.  Hence  the  surface  consists  of 
two  branches  or  sheets  which 
recede  indefinitely  from  the 
FZ-plane  and  from  the  A'-axis. 

In  like  manner  the  sections 
formed  by  all  planes  parallel 
to  the  AF-plane  and  the  ZA-plane  are  hyperbolas  whose  axes 
increase  indefinitely  as  their  planes  recede  from  the  coordi- 
nate planes. 

The  hyperboloid  (1)  is  said  to  ''~  lie  along  the  A'-axis." 

The  equations 


(2) 


a^  ^  P      c"        ' 


y 


u^  ^  <?      ' 


j2  -  ^  -  -L,      -  ^z 

are  the  equations  of  hyperboloids  of  two  sheets  which  lie  along 
the  F-axis  and  the  Z-axis  respectively. 


QUADRIC  SURFACES  323 

li  b  =  c,  c  =  a,  OT  a  =  b,  the  hyperboloids  (1)  and  (2)  are 
hyperboloids  of  revolution. 

It  should  be  noticed  that  the  locus  of  (3),  Ai-t.  125,  is  an  ellljj- 
soid  if  all  the  terms  on  the  left  are  positive,  a  hyperbnlo'id  of 
one  sheet  if  but  one  term  is  negative,  and  a  hyperbolold  of  two 
sheets  if  tivo  terms  are  negative.  If  all  the  terms  on  the  left 
are  negative,  there  is  no  locus.  If  the  locus  is  a  hyperboloid, 
it  will  lie  along  the  axis  corresponding  to  the  term  whose  sign 
differs  from  that  of  the  other  two  terms. 


PROBLEMS 

1.  Discuss  and  construct  the  loci  of  tlie  following  equations  : 

(a)  4x2  + 9?/2^- 1622  =  144.  ^g^  ^x"- -  y"" ->r  Qz"^  =  ZQ. 

(b)  4x2  + 92/2- 16z2  =  144.  (f)  z- -  Ax^  -  ^y"- =  U. 

' (c)  4x2_9^2_i6z2  =  i44.  (g)  16x2  +  ?/2  +  l6z2  =  64. 

(d)  x2  +  16  2/2  +  z2  =  64.  (h)  x2  +  2/-  -  z^  =  25. 

2.  Reduce,  by  translation  of  the  axes,  each  of  the  following  to  a 
standard  form  and  determine  the  type  of  central  quadric  it  represents: 

^.(a)  x2  +  2  2/2  +  2-2  _2x  +  42/ -82 +  10  =  0. 
,    (b)  x2  -  y2  ^_  2  z2  _  6  X  +  2  2/  +  4  2  +  9  =  0. 
^  (c)  2/2  -  x2  _  9  v2  ^  6x  -  2  2/ -  42  +  G  =  0. 

(d)  x2  -  2  2/2  -  4  22  -  2x  -  8  2/  -  8  =  0. 

(e)  4x2 -2/2 -22 -8x -22/  + 0  =  0. 

(f )  4  x2  —  2/2  -  22  —  8  X  —  2  2/  +  4  =  0. 

(g)  3x2  +  4  2/2  -  8  2/  -  z2  =  0.  ( ^ 

3.  Find  the  equations  of  the  planes  whose  intersections  with  the  ellip- 
soid 9x2  +  25 2/2  ^  169 z2  =  1  are  circles.  Ans.    4x  =  ±  12z  +  A:. 

4.  The  square  of  the  distance  of  a  point  from  a  line  is  equal  to  the 
square  of  its  distance  from  a  perpendicular  plane  (a)  increased  by  a  con- 
stant ;  (b)  diminished  by  a  constant.  How  do  the  two  loci  differ  ?  What 
property  have  they  in  common  ? 

5.  A  point  moves  so  that  its  distances  from  a  fixed  point  and  a  fixed 
line  are  in  constant  ratio  n.   Determine  and  name  the  locus 

(a)  when/;i<l.  (c)  when/x=l. 

(b)  when,u>1.  (d)  when  the  point  is  on  the  line. 


/i^ 


324 


NEW  ANALYTIC   GEOMETRY 


6.  A  point  moves  so  that  its  distances  from  a  fixed  point  and  a  fixed 
plane  are  in  a  constant  ratio.  Prove  that  the  locus  is  an  ellipsoid  of  revo- 
lution when  the  ratio  is  less  than  unity,  and  a  hyperboloid  of  revolution 
when  greater  than  unity. 

7.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
two  intersecting  perpendicular  lines  in  .space  is  constant.  Prove  that  the 
locus  is  an  ellipsoid  of  revolution. 


129.  The  elliptic  paraboloid ^  —  =  2cz. 


If  the  coefi&cient 


of  if  \\\  (4),  Art.  125,  is  positive,  the  locus  is  called  an  elliptic 
paraboloid.  A  discussion  of  its  equation  gives  us  the  following 
properties  : 

1.  The  elliptic  paraboloid  is 
symmetrical  with  respect  to  the 
}'Z-plane  and  the  ZA'-plane  and 
the  Z-axis. 

2.  It  passes  through  the  origin, 
but  does  not  intersect  the  axes 
elsewhere. 

3.  Its  traces  on  the  coordi- 
nate planes  are  respectively  the 
conies  o        o 


of  which  the  first  is  a  point-ellipse  and  the  other  two  are 
parabolas. 

4.  The  equation  of  the  curve  in  which  a  plane  parallel  to 
the  A'F-plane,  z  =  k,  cuts  the  paraboloid  is 


|  +  '^=2o., 


+ 


r 


2d-ck      2bhk 


The  curve  is  an  ellipse  if  c  and  k  have  the  same  sign,  but 
there  is  no  locus  if  c  and  k  have  opposite  signs.  Hence,  if  c 
is  positive,  the  surface  lies  entirely  above  the  A'F-plane.  If  A- 
increases  from  0  to  oo,  the  plane  recedes  from  the  A'F-plane 


QUADRIC  SURFACES  325 

and  the  axes  of  the  ellipse  increase  indefinitely.  Hence  the 
surface  recedes  indefinitely  from  the  AT-plane  and  from  the 
Z-axis. 

In  like  manner  the  sections  parallel  to  the  FZ-plane  and  the 
ZA-plane  are  parabolas  whose  vertices  recede  from  the  X  F-plane 
as  their  planes  recede  from  the  coordinate  planes. 

The  paraboloid  is  said  to  "  lie  along  the  Z-axis." 

The  loci  of  the  equations 

(1)  fj  +  |  =  2«a.,     ^;+|'=26y, 

are  elliptic  paraboloids  which  lie  along  the  A'-axis  and  the 
F-axis  respectively. 

It  a  =  b,  the  first  surface  considered  is  a  paraboloid  of  revolu- 
tion whose  axis  is  the  Z-axis  ;  and  if  b  =  c  and  a  =  c,  the  parab- 
oloids (1)  are  surfaces  of  revolution  whose  axes  are  respectively 
the  A'-axis  and  the  F-axis. 

An  elliptic  paraboloid  lies  along  the  axis  corresponding  to 
the  term  of  the  first  degree  in  its  equation,  and  in  the  positive 
or  negative  direction  of  the  axis  according  as  that  term  is 
positive  or  negative. 

130.  The  hyperbolic  paraboloid  —^  —  —  =  1cz.  If  the  coeffi- 
cient of  //"  in  (4),  Art.  125,  is  negative,  the  locus  is  called  a 
hyperbolic  paraboloid. 

1.  The  hyperbolic  paraboloid  is  symmetrical  with  respect  to 
the  FZ-plane  and  the  ZA-plane  and  the  Z-axis. 

2.  It  passes  through  the  origin,  but  does  not  cut  the  axes 
elsewhere. 

3.  Its  traces  on  the  coordinate  planes  are  respectively  the 

-^-•h  =  0,     -  =  2cz,     -%  =  2  cz, 

of  which  the  first  is  a  pair  of  intersecting  lines  and  the  other 
two  are  parabolas. 


32(> 


NEW  ANALYTIC   GEOMETRY 


4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 
A'  }'-plane,  z  =  k,  cuts  the  paraboloid  is 


2  U'ck 


=  1. 


The  locus  is  a  hyperbola.  If  c  is  positive,  the  transverse 
axis  of  the  hyperbola  is  j)arallel  to  the  A'-  or  F-axis  accord- 
ing as  k  is  positive  or  negative.  If  k  increases  from  0  to  oo, 
or  decreases  from  0  to  —  oo,  the 
plane  recedes  from  the  AF-plane 
and  the  axes  of  the  hyperbolas 
increase  indefinitely.  Hence  the 
surface  recedes  indefinitely  from 
the  AF-plane  and  the  Z-axis. 
The  surface  has  approximately 
the  shape  of  a  saddle. 

In  like  manner  the  sections 
I)arallel  to  the  other  coordinate  planes  are  parabolas  whose 
vertices  recede  from  the  AF-plane  as  their  planes  recede  from 
the  coordinate  planes. 

The  surface  is  said  to  "  lie  along  the  Z-axis." 

The  loci  of  the  equations 


^  O    7  V  ^  O 

—  —  Z  01/,     —: :,  =  2  ax, 


are  hyperbolic  paraboloids  lying  along  the  F-axis  and  the  A'-axis 
respectively.  A  hyperbolic  paraboloid  also  lies  along  the  axis 
which  corresponds  to  the  first-degree  term  in  its  equation. 

A  plane  of  symmetry  of  a  quadric  is  called  a  principal  plane. 
Each  paraboloid  has  two  principal  planes  ;  each  central  quadric, 
three.  Axes  of  symmetry  are  called  principal  axes.  A  parab- 
oloid possesses  one  such  axis ;  a  central  quadric,  three.  The 
existence  of  a  center  of  symmetry  for  a  central  quadric  explains 
the  designation  "  central  quadric." 


Plate  II 


Elliptic  Faraboloid  Hyperbolic  Paraboloid 

NONCENTRAL   QUADRICS 


Hyperboloid  of  oue  slict't  Hyperbolic  raiaboloid 

Ruled  Quadrics 


QUADRIC  SURFACES  327 

PROBLEMS 

1.  Discuss  and  construct  the  following  loci : 

(a)  2/2  + z2  =  4  a;.  (e)  9z2  -  4x2  =  288  j^. 

(b)  1/2-  z2  =  4x.  (f)  10x2  +  z^  =  64?/. 

(c)  x2  -  4^2  =  \6y.  (g)  ?/2  _  x2  =  10  2. 

(d)  x~  +  ?/  =  8  z.  (h)  I/'  +  16  z2  +  X  =  0. 

2.  Reduce  by  transformation  of  coordinates  each  of  the  following  to  a 
standard  form  and  determine  the  type  of  paraboloid  it  represents : 

{&)  z  =  xy.  (c)  x2  +  2?/2_  6x  +  4?/  +  3z  +  11  =  0. 

(b)  z  =  x2  +  xy  +  2/2.  (d)  z2-3?/2- 4x  + 2z  — 6?/  + 1  =  0. 

3.  A  point  is  equidistant  from  a  fixed  plane  and  a  fixed  point.  Show 
that  the  locus  is  an  elliptic  paraboloid  of  revolution. 

4.  A  point  is  equidistant  from  two  nonintersecting  perpendicular  lines. 
Show  that  the  locus  is  a  hyperbolic  paraboloid. 

5.  Prove  that  the  parabolas  obtained  by  cutting  (a)  an  elliptic  parabo- 
loid, and  (b)  a  hyperbolic  paraboloid  by  planes  parallel  to  one  of  the 
principal  planes,  are  all  congruent. 

6.  Show  analytically  that  any  plane  parallel  to  the  axis  along  which 
(a)  an  elliptic  paraboloid,  and  (b)  a  hyperbolic  paraboloid  lies,  intersects 
the  surface  in  a  parabola. 

131.  Rectilinear  generators.  The  equation  of  the  hyperboloid 
of  one  sheet,  Art.  127,  may  be  written  in  the  form 

As  this  equation  is  the  result  of  eliminating  k  from  the  equa- 
tions of  the  system  of  lines 

a       c  \         hj''     a       c      k\         h 

the  hyperboloid  is  a  ruled  surface.    Equation  (1)  is  also  the  re- 
sult of  eliminating  k  from  the  equations  of  the  system  of  lines 

a      c  \         b)      a      c      k\         b 

and  the  hyperboloid  may  therefore  be  regarded  in  two  ways  as 
a  ruled  surface. 


328 


NEW  ANALYTIC  GEOMETRY 


In  like  manner  the  hyperbolic  paraboloid  contains  the  two 
systems  of  lines 


a      0 


k' 


and 


ah  ^     a      b        k 


T^ese  lines  are  called  the  rectilinear  generators  of  these 
surfaces.    Hence  the 

Theorem.  Tli.e  liyperboloid  of  one  sheet  and  the  hyperbolic 
paraboloid  have  two  systems  of  rectilinear  generators,  that  is, 
they  may  be  regarded  in  tivo  ivays  as  ruled  surfaces. 

The  two  systems  of  generators  are  shown  in  Plate  11. 


REVIEW  PROBLEMS 


Name  and  draw  the  surfaces 
in  detail  all  their  characteristics 


xy  =  0. 
xy  =  1. 
xy-z. 
xy  =  z^. 
xy  =  z"^  +  1. 
xy  =  z^  +  z. 
x'^  +  y"  —  0. 
X"  +  y"  =  1. 
X-  +  y'^  =  x. 
x2  +  y'^  =  z. 
x^  +  ?/2  =  z^. 
x^  +  y~  =  2  xy. 
x  +  y  =  0. 
x  +  y  =  1. 
x  +  y  =  z. 
X  +  y  =  z^. 
x  +  y  =  xy. 
a;2  +  2/2  =  z2  +  1, 

a;2  +  y-  =  z^  —  1. 
x2  +  2/2  =  1  _  z2. 


8. 


th 

g  following  groups,  giving 

(a) 

x2  +  2/2  =  z-  +  2z. 

(b) 

x2  +  2/2=2:2-22. 

(c) 

x~  +  y-  =  2z-  z2. 

(a) 

x2  +  2  2/2  +  3  z2  =  0. 

(b) 

x2  +  22/2  +  3z2  =  i. 

(c) 

x2  +  2  2/2  +  3  z2  =  2  X. 

(d) 

x2  +  22/2  +  3z2  =  2x 

-1. 

(a) 

x2  +  2  2/2  -  3  z2  =  0. 

(b) 

x2  +  2y2_  322=1. 

(c)  x2  +  2  2/2  — 3z2  =  2a;. 

(d) 

x2  +  22/2-3z2  =  2x 

+  1. 

(e) 

a;2  +  2  7/2-  3^2  =  2x 

-1. 

(f) 

a;2  +  2  2/2  -  3  z2  =  2  X 

-2. 

(a)  xy  +  yz  +  zx  =  0. 

(b) 

z^  +yz  +  zx  =  0. 

(c) 

z  +  yz  -{■  zx  =  0. 

(d) 

Z2  +  X2  +  zx  =  0. 

(e) 

z2  +  x^t+  xy  =  0. 

(f) 

z^  +  xy  +  X  =  0, 

QUADRIC  SURFACES  329 

MISCELLANEOUS  PROBLEMS 

1.  Construct  the  following  surfaces  and  shade  that  part  of  the  first 
intercepted  by  the  second  : 

(a)  a;2  +  42/2  +  9z2  =  36,  x^  +  y^  +  z^  =  16. 

(b)  a;2  +  2/2  +  22  _  64,  x^  +  y^  -8x  =  0. 

(c)  4 X-  +  y"  -  4 z  =  0,  x^  +  4i/^  -  z^  =  0. 

2.  Construct  the  solids  bounded  by  the  surfaces  (a)  x^  +  y^  =  a^. 
z  =  mx,  2  =  0;  (b)  x2  +  ?/2  =  az,  a;2  +  ?/2  =  2  ax,  2  =  0. 

3.  Show  that  two  rectilinear  generators  of  (a)  a  hyperbolic  paraboloid, 
and  (b)  a  hyperboloid  of  one  sheet,  pass  through  each  point  of  the  surface. 

4.  If  a  plane  passes  through  a  rectilinear  generator  of  a  quadric,  show 
that  it  will  also  pass  through  a  second  generator,  and  that  these  generators 
do  not  belong  to  the  same  system. 

5.  The  equation  of  the  hyperboloid  of  one  sheet  may  be  written  in 

?/2         2^  X" 

the  form =  1 By  treating  this  equation  as  In  Art.  131,  we 

obtain  the  equations  of  two  systems  of  lines  on  the  surface.    Show  that 
these  systems  of  lines  are  identical  with  those  already  obtained. 

6.  Show  that  a  quadric  may,  in  general,  be  passed  througii  any  nine 
points. 

7.  If  a  >  6  >  c,  what  is  the  nature  of  the  locus  of 

x2  ?/2  2'- 

+  ,,       ,    + 


a--\      b--\       c2-\ 
if  X  >  a2  ?  if  a2  >  X  >  62  ?  if  62  >  X  >  c2  ?  if  X  <  c2  ? 

8.  Show  that  the  traces  of  the  system  of  quadrics  in  Problem  7  are 
confocal  conies. 

9.  Show  that  every  rectilinear  generator  of  the  hyperbolic  paraboloid 

x2      ?/2  X      y 
—  rr  2cz  is  parallel  to  one  of  the  planes  -  ±     =  0. 

10.  Prove  that  the  projections  of  the  rectilinear  generators  of  (a)  tlic 
hyperboloid  of  one  sheet,  (b)  the  hyperbolic  paraboloid,  on  the  principal 
planes  are  tangent  to  the  traces  of  the  surface  on  those  planes. 

11.  A  plane  passed  through  the  center  and  a  generator  of  a  hyper- 
boloid of  one  sheet  intersects  the  surface  in  a  second  generator  which  is 
parallel  to  the  first. 

12.  Show  how  to  generate  each  of  the  central  (juadrics  by  inoving  an 
ellipse  whose  axes  are  variable. 

13.  Show  how  to  generate  each  of  the  paraboloids  by  moving  a  parabola 


CHAPTER  XX 


EMPIRICAL  EQUATIONS 

132.  A  problem  quite  distinct  from  any  thus  far  treated  in  this 
text  arises  when  it  is  required  to  find  tlte  (yjiidtlon  of  a  curre 
which  shall  pass  throur/h  a  series  of  emplriaxlly  gloen  j^olnts. 
That  is,  we  suppose  that  certain 
values  of  the  vr.riable  and  of 
the  function  are  known  from  an 
actual  experiment,  and  the  cor- 
responding points  are  plotted  on 
cross-section  paper.  A  smooth 
curve  is  then  drawn  to  "  fit " 
these  points,  and  an  equation 
for  this  curve  is  required. 

The  general  treatment  of  this 
important  problem  is  beyond 
the  scope  of  an  elementary  text, 
and  the  following  sections  are  concerned  with  simple  cases  only. 

133.  Straight-line  law.    If  the  curve  suggested  by  the  plotted 
points  is  a  straight  line,  assume  the  law 

(1)  y  =  mx  +  h, 

and  determine  the  values  of  m  and  b  from  the  observed  data. 
The  straight  line  representing  the  required  law  will  not  neces- 
sarily pass  thnnujh  all  the  points  plotted,  for  experimental  work 
is  subject  to  error.  It  is  suflBcient  if  the  line  fits  the  points 
within  the  limits  of  accuracy  of  the  experiment.  In  general, 
the  straight  line  may  be  drawn  through  two  of  the  plotted 
points,  and  m  and  h  may  be  calculated  from  their  coordinates. 

330 


1     t 

"^  *w-. 

^^ . 

V             1 

^ 

"v 

\^ 

rS 

I     \ 

[       \          '    1 

■    1    '-          1        ]_ 

1       \     ' 

\ 

\ 

-        _     -^ 

1  N  M  1 1 — LU — 1  11  M  M  M  1 — >. 

EMPIRICAL  EQUATIONS 


331 


EXAMPLE 

In  an  experiment  with  a  pulley,  the  effort,  E  lb.,  required  to  raise  a 
load  of  Wlb.  was  found  to  be  as  follows : 


w 

10 

20 

30 

40 

50 

60 

70 

80 

00 

100 

E 

n 

61-        11 

9 

10* 

12i 

13| 

15 

16i 

Find  a  straight-line  law  to  fit  these  data. 

Solution.  Plotting  the  points  as  in  the  figure,  it  is  seen  that  the  straight 
line  drawn  through  (30,  6^)  and  (100,  16^)  fits  the  observed  data  very- 
well.   To  find  its  equation,  substitute  these  values  in  the  equation 


(2) 


E  =  jnW+b. 


E 


This   gives   Q,\  =  30  m  +  6,        20 
16^=100?n  +  6.    Solving  for 
m   and   6,    we   find  m  =  /j^g,     ^  15 
b  =  -Y--    Substituting  in  (2),        =| 

(3)        E  =  ^S'-oW+Y,  ^1*^ 


the  required  equation. 

For  the  purpose  of  calcu- 
lation we  write  (3)  in  the  form 

(4)     £■  =  0.146  n"+ 1.86, 


— ~bHi — ' '" 


40  60  80  100     W 

Load  W  in  lbs. 


keeping  three  decimal  places  in  the  coefficient  of  W  in  order  to  secure 
three-figure  accuracy. 

We  now  test  (4)  by  comparing  the  observed  values  and  calculated  values. 


w 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

E,  observed 

3.25 

4.87 

6.25 

7.50 

9 

10.5 

12.2 

13.7 

15 

16.5 

E,  calculated 

3.32 

4.78 

6.24 

7.70 

9.16 

10.6 

12.1 

13.5 

15 

16.5 

The  formula  (4)  may  be  used  for  calculating  values  of  E  for  values  of 
W  intermediate  between  10  and  100,  and  not  given  in  the  table.  For 
example,  the  effort  required  to  raise  a  load  of  25  lb.  is 

E  =  0.146  X  25  -f  1.86  =  5.51  lb. 


332 


NEW  ANALYTIC  GEOMETRY 


PROBLEMS 

The  following  data  treated  in  the  same  way  will  yield  laws  represented 
by  the  formula  y  =  mx  +  b. 

1.  F  is  the  volume  in  cubic  centimeters  of  a  certain  quantity  of 
gas  at  the  temperature  i°  C,  the  pressure  being  constant.  Find  the 
law  connecting  V  and  t. 


t 

27 

33 

40 

55 

68 

V 

109.9 

112.0 

114.7 

120.1 

125. 

Ans.    r=  100  + 0.367  <. 

2.  V  is  the  volume  of  a  certain  quantity  of  mercury  at  a  temperature 
of  $°C.    Find  the  law  connecting  V  and  0. 


e°c. 

18 

36 

60 

72 

90 

F(cc.) 

100.32 

100.65 

101.07 

101.30 

101.61 

Ans.    F=  100  + 0.018(9. 

3.  S  is  the  weight  of  sodium  nitrate  dissolved  by  100  g.  of  water  at  the 
temperature  (°C.    Find  the  law  connecting  S  and  t. 


s 

G8.8 

72.9 

87.5 

102 

t 

-6 

0 

20 

40 

An».    6' =  73.0  +  0. 73  i. 

4.  The  following  are  corresponding  values  of  the  speed  and  induced 
volts  in  an  arc-light  dynamo.  Find  the  law  connecting  volts  and  revolu- 
tions per  miinite. 


Rev.  per  minute,  n 

200 

320 

495 

621 

744 

Volts  induced,  v 

165 

270 

410 

525 

625 

EMPIRICAL  EQUATIONS 


333 


5.  S  is  the  weight  of  potassium  bromide  which  will  dissolve  in  100  s;. 
of  water  at  the  temperature  t°  C.  Find  the  law  connecting  ,S  and  t. 


t 

0 

20 

40 

60 

80 

s 

53.4 

64.6 

74.6 

84.7 

93.6 

6.  Find  the  equation  of  the  straight  lines  that  best  fit  the  following 
data. 


0.5 


0.31 


0.82 


1.5 


1.29 


1.85 


2.5 


2.51 


3.02 


t 

0 

5 

10 

15 

20 

25 

30 

T 

15 

20 

24.4 

28.4 

32 

35.2 

38.2 

2.50 


0.64 


400 


0.80 


500      600 


0.91     0.99 


750 


1.12 


800 


1.15 


900 


1.22 


(a) 


(b) 


(c) 


(d) 


134.  Laws  reduced  to  straight-line  laws.  By  suitable  treat- 
ment of  the  given  data  many  laws  can  be  transformed  into  a 
linear  relation.  Some  cases  of  this  kind  of  frequent  occurrence 
will  now  be  given. 

1.  The  law  V  =  a  -\-  bx\ 

When  the  points  plotted  suggest  a  vertical  parabola  with  its 
vertex  on  the  ?/-axis,  the  assumed  equation  will  have  the  above 
form.  If  now  we  set  x^  =  t,  and  plot  the  values  of  t  and  y,  these 
values  satisfy  the  relation  y  =  a-\- ht,  that  is,  a  straight-line  law. 


w 

3  '  13 

23 

33 

4:] 

F 

f 

1 

13 

91 

n 

334 


NEW  ANALYTIC  GEOMETRY 


EXAMPLE 

An  experiment  to  determine  the  coasting  resistance  R  in  pounds  per  ton 
of  a  motor  wagon  for  the  speed  V  miles  per  hour  gave  the  following  data : 


V 

0 

2i 

5 

'2 

10 

12i 

15 

R 

40 

40 

42 

45 

50 

55 

G3 

Plotting  the  points,  the  curve  suggested   (Fig.  1)   appears  to  be  a 
parabola  with  the  equation 


(1) 


r 

J  00 

lb 

r 

^ 

60 

f 

e 

4 

i*- 

•• 

25 

, 

_ 

_ 

_ 

_ 

_ 

_ 

_ 

R  =  a  +  6F2. 


5  10  15  20 

Speed  V  miles  per  Lour 

Fig.  1 


250    t 


Fig.  2 


To  check  this,  calculate  the  values  of  V^  (table.  Art.  3),  set  V^  =  t,  and 
retabulate  the  data  thus  : 


n=v^) 

0 

H 

25 

56i 

100 

156i 

225 

R 

40 

40 

42 

45 

50 

55 

63 

Plotting  these  points  (Fig.  2),  it  appears  that  they  are  fitted  by  a 
straight  line.  By  the  preceding  section  we  find  the  equation  of  this  line 
to  be  B  =  39.3  +  .107^.    Hence  the  required  law  is 


(2) 


iJ  =  39.3  +  . 107  F2. 


EMPIRICAL  EQUATIONS 


335 


PROBLEMS 

The  following  data  satisfy  laws  of  the  form  y  =  a  +  bx^.    Determine 
the  values  of  a  and  b. 


X 

19 

25 

31 

38   44 

y 

1900 

3230 

4900 

7330 

9780 

Ans.    y  =  5x-  +  102. 


s 

1 

4 

3 
8 

1. 

3 

1 

H 

H 

2 

p 

2 

2| 

H 

5 

n 

lOf 

13| 

221 

3. 


V 

10 

20 

30 

40 

50 

R 

7 

9.1 

14.5 

20 

29 

d 

3 

a 

1 

2 

5 
8 

f 

$ 

1 

S 

663 

1178 

1841 

2651 

3608 

4712 

2.  The  law  y  =  ax". 

Taking  logarithms,  we  have 

(3)  log  i/  =  \og  a  +  n  log  x, 

that  is,  the  logarithms  of  the  given  data  satisfy  a  straight-line 
law.  Hence  in  this  case  tabulate  the  logarithms  of  the  given 
data,  determine  the  straight-line  law  to  lit  them,*  compare 
with  (3)  to  find  a  and  n,  and  substitute  in  y  =  ax". 

This  law,  as  the  study  of  the  problems  on  page  337  will 
show,  has  wide  application. 

^  Logarithmic  squared-paper  is  of  great  convciiiencu  in  tills  I'ase. 


386 


NEW  ANALYTIC   GEOMETRY 


EXAMPLE 

The  following  data  satisfy  a  law  of  the  form  y  —  ax".    Find  the  values 
of  a  and  ?;.  , 

y 


X 

4 

7 

11 

15 

21 

y 

28.G 

79.4 

182 

318 

589 

Solution.    Tabulating  the  values  of  log  r 
and  log  y  (table,  p.  4), 


logjr 

0.602 

0.845 

1.041 

1.176 

1.322 

logy 

1.456 

1.900 

2.26 

2.50 

2.77 

Plotting  x'  =  loga;,  ?/' =  log  ?/,  it  ap- 
pears that  x'  and  y'  satisfy  a  straight-line 
law.    The  equation  is  found  to  be 


- 

- 

- 

- 

- 

- 

- 

- 

v- 

- 

- 

\r\- 

/ 

1 

5 

/ 

n 

} 

J 

I 

I 

\ 

/ 

1 

J 

1 

f 

/ 

1 

1 

1 

1 

i 

(4) 


y'=  .364  +  1.82  x'. 


x'  =  log  X 


Hence,  comparing  with  (3),  n  =  1.82,  log  a  =  .364.    Then  a  =  2.3,  and 
the  required  law  is 


(6) 


?/  =  2.3x1-82.   Ans. 


PROBLEMS 

1.  Find  a  law  of  the  form  y  =  ax»  for  the  following  data: 


X 

2000 

4000 

6000 

8000 

10,000 

y          2869 

8700 

16,660 

26,370 

37,660 

2.  Find  a  law  of  the  form  p  =  od"  to  fit  the  following  data  : 


V 

4 

4.5 

5 

5.5 

6 

7 

p 

110 

97.1 

86.8 

78.4 

71.5 

60.7 

EMPIRICAL  EQUATIONS 


337 


3.  The  time,  t  seconds,  that  it  took  for  water  to  flow  through  a  tri- 
angular notch,  under  a  pressure  of  h  feet,  until  the  same  quantity  was  in 
each  case  discharged,  was  found  by  experiment  to  be  as  follows : 


h 

.043 

.057 

.077 

.094 

.100 

t 

1260 

540 

275 

170 

135 

Find  the  law. 

4.  The  indicated  horse  power  I  required  to  drive  a  vessel  with  a 
displacement  of  D  tons  at  a  ten-knot  speed  is  given  by  the  following  data. 
Find  the  law  connecting  I  and  D. 


D 

1720 

2300 

3200 

4100 

I 

655 

789 

1000 

1164 

5.  u  is  the  volume  in  cubic  feet  of  1  lb.  of  saturated  steam  at  a  pressure 
of  p  lb.  per  square  inch.  Find  the  law  of  the  form  p««  =  const,  connect- 
ing p  and  u. 


u 

26.43 

22.40 

19.08 

16.32 

14.04 

p 

14.7 

17.53 

20.80 

24.54 

28.83 

6.  F  is  the  force  between  two  magnetic  poles  at  a  distance  of  d  centi- 
meters.   Find  the  law  connecting  F  and  d. 


dcm. 

1.2 

1.9 

2.3 

3.2 

4.5 

F dynes 

4.44 

1.77 

1.21 

0.625 

0.316 

7.  1)  is  the  diameter  in  inches  of  wrought-iron  shafting  required  to 
transmit  H  horse  power  when  running  70  revolutions  per  minute. 
Find  a  fornnila. 


H 

10 

20 

30 

40 

50 

60 

70 

80 

D 

2.11 

2.67 

3.04 

3.36 

3.61 

3.82 

4.02 

4.22 

338 


NEW  ANALYTIC  GEOMETRY 


8.  Q  is  the  quantity  of  water  in  cubic  feet  per  second  flowing  tlirougli 
a  riglit-angled  isosceles  notcli  when  the  surface  of  quiet  water  is  H  feet 
above  the  bottom  of  the  notch.    Find  the  law. 


H 

1 

2         3 

4 

Q 

2.63 

15        41 

84.4 

9.  A  certain  ship  draws  h  feet  of  water  and  displaces  V  cubic  feet. 
AYhat  is  the  law  connecting  h  and  F? 


Draft  ft 

18 

13 

11 

9.5 

Displacement  V 

107,200 

65,800 

51,200 

41,100 

135.  Miscellaneous  laws.  In  many  experiments  the  analytic 
form  of  the  law  is  known  in  advance.  If,  however,  such  fact 
is  unknown,  and  if  the  preceding  methods  fail,  the  points  deter- 
mined by  the  data  should  in  any  case  be  plotted,  and  then  the 
shape  of  the  required  curve  may  suggest  an  equation  to  be 
tried.    The  following  problems  furnish  a  variety  in  this  regard. 


PROBLEMS 

1.  The  curve  suggested  in  each  of  the  following  is  a  vertical  parabola 

y  =  a -\- bx -\- cx^ . 
The  values  of  a,  5,  and  c  may  be  found  from  three  pairs  of  values  of 
the  data.    Determine  the  law  in  each  case. 


(a) 


(b)  The  resistance,  R  ohms,  of  a  wire  at  t°  C  is  given  by  the  follow- 
ing table  : 


X 

1 

2 

3 

4 

5         6 

7 

y 

25 

41 

55 

67 

77        85       91 

t 

0 

5 

10 

15 

20 

25 

R 

25 

25.49 

25.98 

26.48 

26.99 

27.51 

EMPIRICAL  EQUATIONS 


339 


(c) 


(d) 


X 

0   0.5 

1 

1.5 

2 

2.5 

3 

y 

5.4   6.3 

6.6   6.1 

5.0 

3.2 

0.6 

u 

0 

20 

40 

60 

80 

100 

V 

2U0 

253 

215 

176 

136 

94 

2.  The  points  may  be  fitted  by  a  branch  of  an  equihiteral  hyperbola 
whose  equation  is  ^ 

In  this  case  plot  y  and  -  —  t^  thus  transforming  into  a  straight-line 
law.    Find  the  law  after  this  manner  for  the  following  data  : 

(a) 


(b) 


X 

4 

5 

6 

7 

8 

y 

4410 

3530 

2940 

2520 

2210 

(c) 


A 

1.06 

2.46 

2.97 

3.45 

3.96 

4.97   5.97 

V 

50.25 

48.7 

47.9 

47.5 

46.8 

45.7 

45 

S 

10 

11 

12 

13 

14 

W 

8370 

48S0 

-  1970 

-490 

-  2600 

3.  In  some  cases  a  branch  of  the  rectangular  hyperbola  {(12),  p.  184), 

xy=bx-\-ay 

a      b 
will  fit  the  points.    Dividing  through  by  xy,  this  becomes  1  =  -  H 

11  ^      ^'^ 

Hence  if  -  =  h,  -  =  v,  are  plotted,  u  and  v  will  satisfy  a  straight-line  law. 
X  y 

Show  tliat  this  is  the  case  in  the  following  and  find  the  law. 


(a) 


X     1  10 

20 

30 

40 

50 

60 

70 

80 

y 

12.8 

17.1 

20 

22.2 

23.1 

23.8 

23.8 

24.2 

340 


NEW  ANALYTIC   GEOMETRY 


(b) 


s 

1 

2 

3 

4 

5 

6 

7 

8 

t 

2.05 

3.23 

3.95 

4.49 

4.87 

5.20 

5.40 

5.60 

4.  Compound-interest  law  (p.  103).    If  the  law  sought  is  of  the  form 
taking  logarithms, 


y  =  ae"^, 
logy  =  loga  +  fcx  loge, 


where  log  e  =  0.434,  as  usual.  Hence  log  y  and  x  satisfy  a  linear  relation, 
and  we  accordingly  plot  x  and  log  y,  determine  the  straight  line,  the 
values  of  a  and  k,  and  substitute. 

Proceed  in  this  manner  in  the  following  : 


(a) 

X 

0 

3.45 

10.85 

19.30 

28.8 

40.1 

53.75 

y 

19.9 

18.9 

16.9 

14.9 

12.9 

10.9 

8.9 

Hint 

Plot  z  and  log  y. 

(b) 

h 

0 

886 

2753 

4763 

6942 

10,693 

P 

30 

29 

27 

25 

23 

20 

Hint.  Plot  h  and  log  p. 
(c) 


t 

0 

10 

27.4 

42.1 

s 

61.5 

62.1 

66.3 

70.3 

X 

0 

2.1 

5.6 

9.3 

11.5 

y 

20 

18.92 

17.34 

15.8 

14.96 

136.  The  problem  under  discussion  requires  for  thorough 
solution  the  method  of  least  squares,  and  for  an  exposition  of 
this  theory  the  student  is  referred  to  treatises  on  that  subject. 


INDEX 


Abscissa,  10 

Algebraic  curve,  44 

Amplitude,  108 

Anchor  ring,  306 

Angle,eccentric,215  ;  vectorial,  119 

Arch,  parabolic,  158 

Area  of  ellipse,  175 

Asymptotes,  51,  170 

Auxiliary  circle,  164 

Axis,  conjugate,  167;  major,  161.; 

minor,  161 ;  transverse,  167 
Axis  of  parabola,  153 

Cardioid,  221 

Catenary,  113 

Center,  instantaneous,  218 

Center  of  conic,  100,  167 

Central  conic,  186 

Central  quadric,  318 

Circle,  point-,  92 

Cissoid  of  Diodes,  54,  210,  220 

Cocked  hat,  55 

Colatitude,  313 

Compound  interest  curve,  103 

Compound  interest  law,  340 

Conchoid  of  Nicomedes,  221 

Confocal  conies,  189 

Conicoid,  316 

Conjugate  diameters,  229 

Coordinates,  oblique,  11 

Cubical  parabola,  46 

Curtate  cycloid,  216 

Cycloid,  208,  212 


Degenerate  ellipse,  179 
Degenerate  hyperbola,  179 
Degenerate  parabola,  179 
Degenerate  quadric,  318 
Director  circle,  225 
Directrix,  153,  186,  295 
Discriminant  of  the  equation  of  a 
circle,  93 

Eccentricity,  162,  168 
Ellipse,  point,  165 
Epicycloid,  217 
Exponential  curves,  102 

Focal  radii  of  conies,  187 
Eocus,  153,  160,  107,  186 
Folium  of  Descartes,  209 
Four-leaved  rose,  126,  223 

Helix,  301 

Hyperbolic  spiral,  132 
Hypocycloid,  217;    of  four  cusps, 
210,  213 ;  of  three  cusps,  206 

Intercepts,  46 
Involute  of  a  circle,  216 

Latus  rectum,  154,  161,  168 
Lemniscate  of  Bernoulli,  55,  122, 

225 
LimaQon  of  Pascal,  55,  222,  225 
Lituus,  132 

Logarithmic  curves,  102 
Longitude,  313 


341 


342 


NEW  ANALYTIC  GEOMETRY 


IMaximum  value  of  a  function,  136 
Minimum  value  of  a  function,  136 

Octant,  232 
Ordinate,  10 

Parabola,  cubical,  46;  semicubical, 

205 
Parabolic  spiral,  223 
Parameter,  8-4 
Period  of  sine  curves,  107 
Point-circle,  92 
Point  of  contact,  191 
Polar  axis,  119 
Pole,  119 

Principal  axes,  326 
Principal  planes,  319,  326 
Probability  curve,  105 
Prolate  cycloid,  216 

Radian,  2 
Radius  vector,  119 


Reciprocal  spiral,  132 
Rose,  three-leaved,  125,  126  ;  four- 
leaved,  126,223;  eight-leaved,  126 

Semicubical  parabola,  205 

Spiral,  hyperbolic  or  reciprocal, 
132  ;  logarithmic  or  equiangular, 
132,  133 ;  parabolic,  223 

Spiral  of  Archimedes,  132 

Strophoid,  54,  226 

Symmetry,  43 

System  of  logarithms,  common, 
101 ;  natural,  101 

Torus,  306 

Traces  of  a  surface,  257 

Triangle  problems,  90 

Vertex  of  a  conic,  153 

Whispering  gallery,  203 
Witch  of  Agnesi,  219 


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